Chem 23.1 Inorganic Analytical Chemistry Laboratory Post-Lab Discussion PDF

## **Chem 23.1** ### Inorganic Analytical Chemistry Laboratory ## **Post-Lab Discussion** ### Experiments 3, 4 and 5 ## **Exam Schedule** ### Second Long Exam - **Date:** October 28, 2024 - **Time:** 1:00 pm to 4:00 pm ### **Coverage:** - Experiment 3 – Acid-Base Titration - Experiment 4 – Precip...

## **Chem 23.1** ### Inorganic Analytical Chemistry Laboratory ## **Post-Lab Discussion** ### Experiments 3, 4 and 5 ## **Exam Schedule** ### Second Long Exam - **Date:** October 28, 2024 - **Time:** 1:00 pm to 4:00 pm ### **Coverage:** - Experiment 3 – Acid-Base Titration - Experiment 4 – Precipitation Titration - Experiment 5 – Complexometric Titration ## **Module 5: Acid-Base Titration** ## **Learning Outcomes** At the end of this activity, you must be able to: - Prepare HCl and NaOH solution - Standardize NaOH against a primary standard KHP - Determine the concentration of acetic acid in vinegar ## **Introduction to Titrimetry** - Titration methods are based on determining the quantity of a reagent of known concentration (titrant) that is required to react completely with the analyte (titrand). - The reagent may be a standard solution of a chemical or an electric current of known magnitude. - In volumetric titrations, the volume of a standard reagent is the measured quantity. - In coulometric titrations, the quantity of charge required to complete a reaction with the analyte is the measured quantity ## **Important Terminologies** - **Standard solution (standard titrant):** a reagent of known concentration that is used to carry out a volumetric titration. - **Equivalence point:** a theoretical point reached when the amount of added titrant is chemically equivalent to the amount of analyte in the sample. - **End point:** the point in a titration when a physical change occurs that is associated with the condition of chemical equivalence. - **Indicators:** added to the analyte solution to produce an observable physical change (signaling the end point) at or near the equivalence point ## **Indicators** A table with the following data is shown in the image: | Indicator | pH range for color change | |---|---| | Methyl violet | 0 -2 - Yellow to Violet - 14 | | Thymol blue | 0 -2 - Red to Yellow - 8 | | | 8 - 14 - Yellow to Blue | | Methyl orange | 3 - 4 - Red to Yellow | | Methyl red | 4 - 6 - Red to Yellow | | Bromthymol blue | 6 - 8 - Yellow to Blue | | Phenolphthalein | 8 - 10 - Colorless to Pink | | Alizarin yellow R | 10 - 12 - Yellow to Red | The image also contain two illustrations. One is on the pH ranges for common acid-base indicators. The other one shows the solutions containing three common acid-base indicators at various pH values. ## **Titration** An illustration is shown that depicts the procedure for titrating an acid against a standard solution of NaOH in a burette, where a few drops of acid-base indicator, phenolphthalein, are added. Phenolphthalein is colorless in acidic solution but takes on a pink color in basic solution. ## **Titration** An illustration is shown that depicts the procedure for determining the concentration of a solution from titration with a standard solution: - **Volume of standard solution needed to reach equivalence point** - **Use of molarity of standard solution** - **Moles of solute in standard solution** **Use coefficients from balanced equation** - **Concentration (molarity) of unknown solution** - **Use volume of unknown solution** - **Moles of solute in unknown solution** ## **Sample Problem** The distinctive odor of vinegar is due to acetic acid, $CH_3COOH$, which reacts with sodium hydroxide in the following fashion: $$CH_3COOH (aq) + NaOH (aq) \rightarrow CH_3COONa (aq) + H_2O (l)$$ If 20 mL of vinegar needs 14.5 mL of 1.0 M NaOH to reach the equivalence point in a titration. - What is the concentration of the acid solution? - How many grams of acetic acid are present in vinegar? **Solution:** - **conc:** $$14.5mL NaOH \times \frac{1L}{1000mL}\times \frac{1.0 mol NaOH}{1L NaOH} \times \frac{1 mol Acetic acid}{1 mol NaOH} = 0.725 M$$ - **g Ac:** $$14.5mL NaOH \times \frac{1L}{1000mL}\times \frac{1.0 mol NaOH}{1L NaOH} \times \frac{1 mol Acetic acid}{1 mol NaOH} \times \frac{60.06 g Ac}{1 mol Ac} = 0.871 g$$ ## **Acids and Bases** - Recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HCIO3, and HCIO4 - For monoprotic strong acids, $[H_3O+]=[acid]$. - Strong bases are the soluble hydroxides, which are the alkali metal: NaOH, KOH, and heavier alkaline earth metal hydroxides: Ca(OH)2, Sr(OH)2, and Ba(OH)2. An illustration is shown that depicts the relative acid and base strength of different acids and bases. ## **pH** - pH is defined as the negative (base-10) logarithm of the hydronium ion concentration. - $$pH = -log [H_3O+]$$ A table is shown that depicts the solution type, hydrogen ion concentration, hydroxide ion concentration and their pH scale. ## **Autoionization of Water** - Water is amphoteric (it can act as a base or acid). - $$H_2O (l) + H_2O (l) \rightleftharpoons OH(aq) + H_3O+ (aq)$$ - The equilibrium expression for this process is: $$K_p = \frac{[H_3O+][OH-]}{[H_2O]^2}\rightarrow K_w=[H_3O+][OH]$$ - This special equilibrium constant is referred to as the ion-product constant for water, $K_w$. $$K_w = 1.0 \times10^{-14} (at 25^\circ C)$$ - In pure water, $$K_w=[H_3O+][OH-]=1.0 \times 10^{-14}$$ Because in pure water, $[H_3O+]=[OH-]$, $$[H_3O+]=1.0 \times 10^{-7}$$ ## **pOH and other "p" scales** - The "p" in pH tells us to take the negative log of the quantity (in this case, hydronium ion). - $$pOH = -log [OH-]$$ - Since it has been known that: $$[H_3O+][OH-]=K_w=1.0 \times 10^{-14}$$ - $$-log[H_3O+]+ -log[OH-]=-logK_w=14.00$$ Therefore, - $$pH+ pOH = pK_w=14.00$$ ## **Sample Exercise** A research chemist adds a measured amount of HCl gas to pure water at 25C and obtains a solution with $[H_3O+]=3.4 \times 10^{-4} M$. - Calculate $[OH-]$. - Is the solution neutral, acidic, or basic? **Solution:** - Calculating $[OH-]$: $$K_w=[H_3O+][OH-]=1.0\times 10^{-14}$$ $$[OH-]=\frac {1.0 \times 10^{-14}}{3.4 \times 10^{-4}} = 2.9 \times 10^{-11}$$ - Calculating pH : $$pH=-log[3.4 \times 10^{-14}]=3.47$$ Therefore, the solution is acidic. ## **Titrating Strong Acids and Strong Bases** Let's consider the titration of 50.0 mL of 0.100 M HCl using a titrant of 0.200 M NaOH. When a strong base and a strong acid react the only reaction of importance is: $$H_3O+(aq) + OH®(aq) \rightarrow 2H_2O(l)$$ Calculating the volume of NaOH needed to reach the equivalence point, $V_{eq}$: $$moles HCl = moles NaOH $$ $$M_a \times V_a = M_b \times V_b$$ $$V_{eq} = V_b = \frac{M_aV_a}{M_b} = \frac {(0.100 M)(50.0 mL)}{(0.200 M)}= 25.0 mL$$ ### **Before the equivalence point** HCl is present in excess and the concentration of unreacted HCl determines the pH. At the start of the titration, the solution is 0.100 M in HCl, which, because HCl is a strong acid, means the pH is: $$pH=-log[H_3O+]=-log[HCI]=-log(0.100)=1.00$$ ### **After adding 10.0 mL of NaOH** $$[HCI] = \frac {(mol HCl)initial-(mol NaOH) added}{total volume}=\frac{ MaVa - MbVb}{Va + Vb}$$ $$[HCI]=\frac {(0.100 M)(50.0 mL)-(0.200 M)(10.0mL)} {50.0mL + 10.0mL}=0.0500M$$ $$pH= -log(0.0500) M = 1.300$$ ### **At the equivalence point (25mL NaOH is added)** The moles of HCl and the moles of NaOH are equal. Since neither the acid nor the base is in excess, the pH is determined by the dissociation of water. $$K_w = 1.00 × 10-14 = [H_3O+][OH®] = [H_3O+]^2$$ $$[H_3O+]=1.00 × 10^-7$$ $$pH = -log(1.00 x 10^-7) M = 7.000$$ ### **For volumes of NaOH greater than the equivalence point (30mL NaOH is added)** For volumes of NaOH greater than the equivalence point, the pH is determined by the concentration of excess OH−. $$[OH®]= \frac{(mol NaOH)added-(mol HCl)initial}{total volume} = \frac{ MbVb-MaVa}{Va+Vb}$$ $$[OH®]=\frac{(0.200 M)(30.0 mL)-(0.100 M)(50.0 mL)} {30.0mL+50.0mL}=0.0125 M$$ To find the pH: $$[H_3O+]=\frac{K_w}{[OH-]}=\frac{1.00 \times 10^{-14}}{0.0125}=8.00 \times 10^{-13} M$$ $$pH = -log(8.00 x 10-13) M = 12.100$$ or $$pOH=-log(0.0125)=1.900$$ $$pH=pK_w-pOH = 14.000 - 1.900 = 12.100$$ ### **Titration curve for the titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOH** A table and an illustration depict the titration curve for the titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOH, where the red points correspond to the data in the table and the blue line shows the complete titration curve. ## **Titration curve for SA-SB** An illustration is shown that depicts the titration curve for strong acid-strong base titration. It shows the following: - **At the beginning:** the solution contains SA only, and the pH is calculated from the concentration of that solution. - **At pre-equivalence point:** The added titrant will consume H+, reducing its concentration in the solution. The pH is calculated from the unreacted H+ that remains in the solution. - **At the equivalence point:** Complete neutralization of acid/base. The solution contains only the product of SA-SB neutralization: water and salt. pH is that of pH of water= 7. - **At post-equivalence point:** The pH is governed largely by the concentration of the excess titrant. ## **Titrating a weak acid with a strong base** Let's consider the titration of 50.0 mL of 0.100 M acetic acid, $CH_3COOH$, with 0.200 M NaOH. Calculating the volume of NaOH needed to reach the equivalence point, $V_{eq}$: $$mol CH_3COOH = mol NaOH$$ $$M_a \times V_a = M_b \times V_b$$ $$V_{eq} = V_b = \frac {MaVa}{Mb} = \frac {(0.100 M)(50.0 mL)}{(0.200 M)} = 25.0 mL$$ Since acetic acid is a weak acid, we calculate the pH by: $$CH_3COOH(aq)+ H_2O(l) \rightleftharpoons H_3O+(aq) + CH_3COO¯ (aq)$$ $$K a = \frac {[H_3O+][CH_3COO-]}{ [CH_3COOH]} = \frac{(x)(x)}{0.100 - x} = 1.75 \times 10^{-5}$$ $$pH=-log(x) M = 2.88$$ Adding NaOH converts a portion of the acetic acid to its conjugate base, $CH_3COO-$. $$CH_3COOH(aq)+ OH_-(aq) \rightarrow H_2O(l)+ CH_3COO_-(aq)$$ **Henderson-Hasselbalch equation:** $$pH=pK_a+log[\frac{A^-}{HA}]$$ ### **Before the equivalence point:** **The concentration of unreacted acetic acid is:** $$[CH_3COOH] = \frac{(mol CH_3COOH)initial -(mol NaOH)added}{total volume}=\frac{ MaVa - MbVb}{Va + Vb}$$ **And the concentration of acetate is:** $$[CH_3COO-]= \frac{(mol NaOH)added}{total volume} = \frac{MbVb}{Va + Vb}$$ ### **After adding 10.0 mL NaOH:** $$[CH_3COOH] = \frac{(0.100 M)(50.0 mL)-(0.200 M)(10.0mL)} {50.0mL + 10.0mL}=0.0500M$$ $$[CH_3COO-]= \frac{(0.200 M)(10.0 mL)}{ 50.0 mL + 10.0 mL} = 0.0333 M$$ $$pH= 4.76 + log \frac{0.0333 M}{0.0500 M} = 4.58$$ ### **At the equivalence point:** (After adding 25.0 mL NaOH) $$[CH_3COO-]= \frac{(mol NaOH)added}{total volume}=\frac{(0.200 M)(25.0 mL)}{50.0 mL + 25.0 mL}=0.0667 M$$ **The pH of the weak base:** $$CH_3COO¯(aq)+ H_2O(l) \rightleftharpoons OH® (aq) + CH_3COOH(aq)$$ $$K_b=\frac{[OH^-][CH_3COOH]}{[CH_3COO-]}=\frac{(x)(x)}{0.0667-x}=5.71 \times 10^{-10}$$ $$x = [OH_ ] = 6.17 × 10-6 M$$ $$[H_3O+]= \frac{K_w}{[OH^-]} =\frac{1.00 × 10^{-14}}{6.17 × 10^{-6}} = 1.62 × 10^{-9} M$$ $$pH = -log(1.62 x 10^{-9}) M = 8.790$$ ### **After the equivalence point:** **After adding 30.0 mL NaOH:** $$[OH®]= \frac{(0.200 M)(30.0 mL)-(0.100 M)(50.0 mL)} {30.0mL+50.0mL}=0.0125 M$$ $$K_w=[H_3O+][OH-]$$ $$[H_3O+]= \frac{K_w}{[OH^-]} =\frac{1.00 × 10^{-14}}{0.0125}=8.00 \times 10^{-13} M$$ $$pH = -log(8.00 x 10^{-13}) M = 12.097$$ ### **Titration curve for the titration of 50.0 mL of 0.100 M Acetic Acid with 0.200 M NaOH** A table and an illustration depict the titration curve for the titration of 50.0 mL of 0.100 M Acetic Acid with 0.200 M NaOH, where the red points correspond to the data in the table and the blue line shows the complete titration curve. ## **Titration curve for WA-SB** An illustration is shown that depicts the titration curve for weak acid-strong base titration. It shows the following: - **At the beginning:** the solution contains only a weak acid or a weak base, and the pH is calculated from the concentration of that solute and its dissociation constant. - **At pre-equivalence point:** A buffer solution is formed. The pH of the buffer can be calculated from the analytical concentrations of the conjugate base and the concentrations of the weak acid that remains. - **At the equivalence point:** the solution contains only the conjugate of the weak acid being titrated (that is, a salt). The pH is calculated from the concentration of this product. - **At post-equivalence point:** The pH is governed largely by the concentration of the excess titrant. ## **Experiment 3: Determination of Acetic Acid Content in Vinegar** ## **General Procedure** - **Standardization of standard solution:** Determine the concentration of standard solution by titrating against a primary standard. - **Titration of sample:** Titrate sample with standardized solution. - **Data Analysis:** Calculate the concentration of analyte in sample using stoichiometry and dimensional analysis based on the volume of titrant used. ## **Calculation of Predicted Volume of NaOH** A table is shown that depicts the theoretical volume of NaOH consumed for each trial: | Trial | Actual Weight | Corrected Wt. KHP | Molarity NaOH | Predicted Vol. NaOH | |---|---|---|---|---| | 1 | 0.8186 g | 0.8178 g | 0.10 M | 40.00 mL | | 2 | 0.8067 g | 0.8059 g | 0.10 M | 39.50 mL | | 3 | 0.8259 g | 0.8251 g | 0.10 M | 40.40 mL | **Dimensional analysis:** $$M NaoH = \frac {mol NaOH}{L}$$ $$L NaoH = \frac {mol NaOH}{M NaOH}$$ $$L NaoH = wt KHP(g) \times \frac {1 mol KHP}{204.22 g KHP} \times \frac{1 mol NaOH}{1 mol KHP}$$ $$L NaoH = \frac{wt KHP (g) \times \frac {1 mol KHP}{204.22 g KHP} \times \frac{1 mol NaOH}{1 mol KHP}}{M NaOH}$$ **Example:** For trial 1: $$L NaoH = \frac{0.8178 g KHP \times \frac {1 mol KHP}{204.22 g KHP} \times \frac{1 mol NaOH}{1 mol KHP}}{0.1 M NaOH} = 0.040044 L = 40.0 mL$$ **Notice how the units cancel out, giving you the desired unit. If you arrived with a different unit using dimensional analysis, the answer probably isn't correct.** ## **Standardization of NaOH Solution** A table is shown that depicts the average molarity of NaOH Standard Solution. | Trial | Actual Weight | Corrected Wt. KHP | Vol. NaOH Used | M NaOH | |---|---|---|---|---| | 1 | 0.8186 g | 0.8178 g | 39.42 mL | 0.1016 M | | 2 | 0.8067 g | 0.8059 g | 39.19 mL | 0.1007 M | | 3 | 0.8259 g | 0.8251 g | 39.92 mL | 0.1012 M | **Average Molarity** = 0.1012 **Dimensional analysis:** $$M NaoH = \frac {mol NaOH}{L NaOH} = \frac{wt. g KHP \times \frac {1 mol KHP}{204.22 g KHP} \times \frac{1 mol NaOH}{1 mol KHP}}{L NaOH}$$ **Sample calculation:** For trial 1: $$M NaoH = \frac {mol NaOH}{L NaOH} = \frac{0.8178. g KHP \times \frac {1 mol KHP}{204.22 g KHP} \times \frac{1 mol NaOH}{1 mol KHP}}{0.03942 L NaOH} = 0.1016 M$$ ## **Determination of %HOAc (w/v)** A table is shown that depicts the %HOAc content of vinegar sample: | Trial | Vol. sample (mL) | Vol NaOH (mL) | % HOAC (w/v) | |---|---|---|---| | 1 | 25.0 | 39.42 | 4.152% | | 2 | 25.0 | 39.19 | 4.1465% | | 3 | 25.0 | 39.92 | 4.154% | | **Average %HOAc (w/v)** | **4.151%** | | **Standard deviation** | **0.004** | **Dimensional analysis:** $$% HOAc (w/v) = \frac{g HOAc}{mL Sample} \times 100% $$ $$% HOAc (w/v) = \frac{0.1012 mol NaOH}{L NaOH} \times L NaOH \times \frac{1 mol HOAc}{1 mol NaOH} \times \frac{60.052 g HOAc}{mol HOAc} \times \frac{250}{50} \times 100%$$ *You are only analyzing 50 mL of the 250 mL sample, or a fifth of the sample. Multiplying by a factor of 5 gives you the total amount of HOAc in the sample.* **Sample calculation:** For trial 1: $$% HOAc (w/v) = \frac{0.1016 mol NaOH}{L NaOH} \times 0.03942 L NaOH \times \frac{1 mol HOAc}{1 mol NaOH} \times \frac{60.052 g HOAc}{mol HOAc} \times \frac{250}{50} \times 100% = 4.15%$$ ## **Questions for Discussion** - Why is there a need to standardize the NaOH solution? - Why is there a need to use CO2-free water in this analysis? - Search literature and discuss the principle behind why phenolphthalein changes color. - Are ‘equivalence point' and ‘end point' the same? If yes, why? Otherwise, differentiate the two. - Aside from repeating the experiment, search literature and suggest what can be done to remedy a sample that is ‘over-titrated'. - If the manufacturer claims 4.5% acid content, evaluate the results you've obtained. ## **Experiment 3b: Strong Acid-Strong Base Titration Curve** ## **Detection of End-Point** From the titration curve, the end point could be taken as the inflection point (point of a curve at which a change in the direction of curvature occurs) in the steeply rising portion of the curve. The image displays three illustrations. One is of the titration cure, the second one shows the first derivative of the titration curve and the third one is the second derivative of the titration curve. ## **Results: Experimental Titration Curve** A table and an illustration depict the HCI-NaOH titration curve. The data in the table shows the volume of titrant and pH. ## **Results: 1st Derivative** A table and an illustration depict the 1st derivative of titration curve. The data in the table shows the average volume of titrant and the change in pH/ change in volume respectively. ## **Results: 2nd Derivative** A table and an illustration depict the 2nd derivative of titration curve. The data in the table shows the average volume of titrant and the change in 2nd derivative of pH/ volume respectively. ## **Results: Theoretical Titration Curve** An illustration is shown that depicts the theoretical titration curve for the titration of a strong acid with a strong base. ## **Module 6: Precipitation Titration** ## **Learning Outcomes** At the end of this activity, you must be able to: - Perform precipitation titration using an argentometric method - Standardize silver nitrate solution against a primary standard NaCl - Determine the chloride concentration in a solid sample using the Mohr method. ## **Recall** - **Precipitation reaction:** occurs when two ionic compounds are combined to form two new ones by exchanging cations and anions with one another. One of the resulting compounds is soluble in water while the other is insoluble hence it is precipitated out of the solution. To know which compound will precipitate, the rules on solubility must be considered. - **Ex:** $Na2CO3 (aq) + CaCl2 (aq) \rightarrow 2NaCl (aq) + CaCO3 (s)$ ## **Precipitation reaction** The image displays an illustration that depicts a precipitation reaction where potassium iodide and lead nitrate react to form lead iodide. The illustration shows a solution of potassium iodide and lead nitrate before reacting and after forming a precipitate. Reactions that result in the formation of an insoluble product are called precipitation reactions. A precipitate is an insoluble solid formed by a reaction in solution ## **Solubility** A table is shown showing the solubility guidelines for common ionic compounds in water and exceptions to these guidelines. It lists soluble ionic compounds containing NO3-, CH3COO-, Cl, Br and SO42- and their exceptions respectively. It also lists insoluble ionic compounds containing S2-, CO32-, PO43- and OH- and their exceptions respectively. The **solubility** of a substance at a given temperature is the amount of the substance that can be dissolved in a given quantity of solvent at the given temperature. ## **Sample Exercise** - Classify the following ionic compounds as soluble or insoluble in water: - Sodium carbonate - Lead sulfate - Barium nitrate - Ammonium phosphate - Will a precipitate form when solutions of AgNO3 and NaCl are mixed? ## **Precipitation Titration** - Precipitation titrations are based on reactions that yield ionic compounds of limited solubility. Precipitation titrimetry is one of the oldest analytical techniques, dating back to the mid-1800s. The slow rate at which most precipitates form, however, limits the number of precipitating agents that can be used in titrations to a handful. ## **Precipitation Titration** A table is shown that shows the typical inorganic Complex-Forming Titrations by listing the titrant, analyte, and remarks for these titrations. Titrations involving the use of silver nitrate as precipitating agent are collectively called **argentometric titrations**, which are mostly used methods in determining chloride and other several anions. ## **Argentometric Titrations** A table is shown that depicts the method, titrant, indicator and observation for AgNO3 titrations. The methods are Mohr, Volhard and Fajans. ## **Mohr Method ** - The **Mohr Method**, named after Karl Friedrich Mohr, uses potassium chromate (K2CrO4) as indicator against the standard silver nitrate (AgNO3) in the determination of chloride, bromide and cyanide ions. Silver ions react with chromate to form the brick-red silver chromate (Ag2CrO4) precipitate in the equivalence-point region, signaling the end point. - An illustration shows the reaction between chloride and silver ions forming a white precipitate and chromate and silver ions forming a brick-red precipitate in the equivalence point region. **Determination of chloride using Mohr Method** - **Eqn 1:** $$cl-+Ag+ \rightarrow AgCl(s)$$ - **Eqn 2:** $$Cro_{4}^{2-} + Ag+ \rightarrow Ag_2CrO_4(s)$$ ## **Sample Exercise** A 50.00 mL sample was subjected to Mohr titration for chloride determination. A few drops of K2CrO4 indicator were added to the sample. Then the sample consumed 15.70 mL of 0.0800 M AgNO3 solution. Calculate the amount of chloride (mg/L) in the sample. **Solution:** $$Ag^+ + cl^- \rightarrow AgCl(s)$$ - Calculating mol of $Ag^+$ and $Cl^-$ : $$mol Ag^+=mol Cl^-= 0.01570 L \times 0.0800 \frac{mol}{L} = 1.256 \times 10^{-3} $$ - Calculating $mg Cl/L$: $$\frac{mg Cl}{L}=\frac{1.256 \times 10^{-3} mol Cl^- \times \frac {35.45 g}{1mol Cl} \times \frac{1000 mg}{1 g}}{0.05000 L} =\frac{mgCl}{L} =891$$ - Calculating $mg Cl/L$ in terms of $mgl$: $$891 \frac{mg Cl}{L}$$ ## **Experiment 4: Determination of Chloride in the Sample** ## **General Procedure** - **Standardization of AgNO3** : Determine the concentration of standard solution by titrating against a primary standard. - **Titration of sample:** Titrate sample with standardized solution. - **Data Analysis:** Calculate the concentration of analyte in sample using stoichiometry and dimensional analysis based on the volume of titrant used. ## **Standardization of AgNO3** A table is shown that depicts the sample data for determination of molarity of silver nitrate: | Trial | mass Nacl (g) | mL AgNO3 used | Molarity AgNO3 | |---|---|---|---| | 1 | 0.1016 | 18.50 | 0.0940 | | 2 | 0.1027 | 18.40 | 0.0955 | | 3 | 0.1001 | 18.50 | 0.0926 | **Average Molarity EDTA** = 0.0940 M ## **Molarity of AgNO3** **Dimensional analysis:** $$M of AgNO_3 = \frac{weight of NaCl \times Formula mass of NaCl \times \frac{1 mol AgNO_3}{1 mol NaCl}}{\text{Corrected volume of } AgNO_3 \times \frac{1 L}{1000 mL}}$$ **For Trial 1:** $$M of AgNO_3 =\frac{(0.1016 g NaCl)(\frac{1 mol NaCl}{58.44 g NaCl})(\frac{1 mol AgNO_3}{1 mol NaCl})} {18.5 mL AgNO_3 \times \frac{1 L}{1000 mL}}$$ $$M of AgNO_3 = 0.0940 M$$ ## **Standardization of AgNO3** A table is shown that depicts the sample data for determination of molarity of silver nitrate. | Trial | mass of sample (g) | nL AgNO3 used | mg Cl | % cl | % NaCl | |---|---|---|---|---|---| | 1 | 0.0507 | 9.40 | 31.3179 | 61.77 | 101.8307 | | 2 | 0.0503 | 9.40 | 31.8291 | 62.28 | 104.3159 | | 3 | 0.0508 | 9.20 | 30.1989 | 58.45 | 97.9989 | **Average** | **31.1153** | **60.83** | **101.3818** | ## **mg of Cl** **Dimentional analysis:** At equivalence point, $$moles NaCl = moles AgNO_3$$ $$moles NaCl = M_{AgNO_3} \times corrected V_{AgNO_3}$$ $$mg Cl = mol NaCl \times \frac{1 mol Cl^-}{1 mol NaCl}\times Formula Mass of Cl \times \frac{1000 mg}{1 g}$$ $$mg Cl = M_{AgNO_3} \times corrected V_{AgNO_3} \times \frac{1 mol Cl^-}{1 mol NaCl} \times Formula Mass of Cl \times \frac{1000 mg}{1 g}$$ **For Trial 1:** $$mg Cl = 0.0940 \frac{mol AgNO_3}{L} \times 0.0094 L AgNO_3 \times \frac{1 mol NaCl}{1 mol AgNO_3} \times \frac{1 mol Cl^-}{1 mol NaCl} \times \frac{35.45 g Cl^-}{mol Cl^-} \times \frac{1000 mg}{1 g}$$ $$mg Cl = 31.3179 M$$ ## **% Cl** **Dimensional analysis:** $$%Cl=\frac {mg Cl}{mg sample}\times 100$$ **For Trial 1:** $$%Cl=\frac {31.3179 mg}{50.7 mg} \times 100$$ $$%Cl= 61.77%$$ ## **% NaCl** **Dimensional analysis:** $$%NaCl = \frac{mg Cl \times \frac{1 g}{1000 mg} \times Formula Mass of Cl \times \frac{1 mol NaCl}{1 mol Cl} \times Formula Mass of NaCl}{g sample} \times 100%$$ **For Trial 1:** $$%NaCl = \frac{31.3179 mg Cl \times \frac{1 g