pH and Indicators PDF

**pH and Indicators** **Self-Ionisation of Water** Water only conducts electricity when it contains ions dissolved in it. However, even pure water conducts a very small current, so there must be some ions present. The ions exist because water self-ionises:\ H~2~O ↔ H^+^ + OH^-^\ This is referred t...

**pH and Indicators** **Self-Ionisation of Water** Water only conducts electricity when it contains ions dissolved in it. However, even pure water conducts a very small current, so there must be some ions present. The ions exist because water self-ionises:\ H~2~O ↔ H^+^ + OH^-^\ This is referred to as the self-ionisation of water. The concentration of H^+^ ions and OH^-^ ions in pure water is extremely small, i.e., the equilibrium is very much on the left-hand side of the equation. As the number of H~2~O molecules that self-ionise is so small, we can say that the concentration of H~2~O stays constant. Equilibrium constant may be written as\ K~c~ = \[H^+^\]\[OH^-^\]\ \[H~2~O\] Because any variation is term \[H~2~O\] is so small in relation to size, the above equation can be rearranged to give\ K~w~ = K~c~\[H~2~O\] = \[H^+^\]\[OH^-^\] K~w~ = Ionic product of water For water at 25^o^C it has been found that K~w~ is 1x10^-14^.\ This means 𝐾~𝑤~ = \[𝐻 ^+^\]\[𝑂𝐻^−^\] = 1 × 10^−14^\ To find \[H^+^\], we need to use the fact that \[H^+^\]=\[OH^-^\]\ 𝐾~𝑤~ = \[𝐻 ^+^\]\[𝐻 ^+^\] = 1 × 10^−14^\ 𝐾𝑤 = \[𝐻 ^+^\]^2^ = 1 × 10^−14\ ^\[𝐻 ^+^\] = [\$\\sqrt{1\\ \\times \\ 10\^{- 14}\\ }\$]{.math.inline}\ \[𝐻 ^+^\] = 1 × 10^−7^ **Measurement of Acidity -- The pH Scale** \*\*Definition: pH = -log~10~\[H^+^\] where the square brackets indicate concentration in moles per litre.\*\*\ The pH of a solution is the negative logarithm to the base 10 of the hydrogen ion concentration measured in moles per litre.\*\* A chart of acid content Description automatically generated(1) \- The pH of an acidic solution is less that 7\ - The pH of a basic(alkaline) solution is greater than 7\ - The pH of a neutral solution is equal to 7 Calculate the pH of a solution whose hydrogen ion concentration is 7.1 x 10^-7^ mol/L. pH = -log~10~\[H^+^\]\ pH = -log~10~(7.1 x 10^-7^) ph = 6.15 Calculate the hydrogen ion concentration (moles per litre) of a solution whose pH is 3.7 pH = -log~10~\[H^+^\] = 3.7\ \[H^+^\] = antilog (3.7)\ \[H^+^\] = 1.995 x 10^-4^ mol/L Instruments that measure pH \- pH data loggers are the most accurate method of measuring pH.\ - Universal indicator paper by comparison to colour chart **Strengths of Acids and Bases** \*\*Definition: A strong acid is a good proton donor.\*\* e.g., HCl, H~2~SO~4~ \*\*Definition: A weak acid is a poor proton donor.\*\* e.g., Ethanoic acid In order to represent the strength of an acid, chemists use the symbol K~a~ which is the dissociation constant and is defined as\ K~a~ = \[H^+^\]\[A^-^\]\ \[HA\] Acid dissociation constants derive from equilibrium constants for particular acids by assuming that \[H~2~O\] is constant. Strong acids have large values of K~a~ and weak acids have small values of K~a~. \*\*Definition: A strong base is a good proton acceptor\*\* e.g., NaOH, KOH\ \*\*Definition: A weak base is a poor proton acceptor\*\* e.g., NH~3~ In order to represent the strength of a base, chemists use the symbol K~b~ which is the dissociation constant and is defined as\ K~b~ = \[NH~4~^+^\]\[OH^-^\]\ \[NH~3~\] **Relative Strengths of Acids and Bases** HCl + H~2~O H~3~O^+^ + Cl^-^\ strong acid weak base Conjugate acid-base pair Since HCl is a strong acid, the reaction goes almost completely from left to right. Since the Cl^-^ ion is a poor proton acceptor, we say it is a weak conjugate base. A strong acid has a weak conjugate base. CH~3~COOH + H~2~O ↔ CH~3~COO + H~3~O^+\ ^weak acid strong base Conjugate acid-base pair\ Since the equilibrium lies on the left-hand side, this implies that the CH~3~COO^-^ is a good proton acceptor so it is a strong base. A weak acid has a strong conjugate base. **Calculating the pH of Strong Acids and Strong Bases** Strong Acids and Bases can be assumed to be fully dissociated in water. Calculate the pH of a 0.04 M H~2~SO~4~ solution H~2~SO~4~ 2H^+^ + SO~4~^2-^ One mole of H~2~SO~4~ gives two moles of H^+^ in solution 0.4 mole 0.8 mole pH = -log~10~\[H^+^\] = -log~10~(0.08) = 0.097 pH = 0.097 pH of a strong base: pOH = -log~10~\[OH^-^\] Calculate the pH of a 0.15 M NaOH solution 1 mole NaOH 1 mole OH^-^ pOH = -log~10~\[OH^-^\] = -log~10~(0.15) = 0.82 pH = 14 -- pOH = 14 -- 0.82 = 13.18 pH = 13.18 **Limitations of the pH Scale** \- Limited to the 0.1 - 14 range even though, theoretically, pH values are possible outside this range.\ - The pH scale is limited to dilute aqueous solutions at 25°C\ - The pH scale does not work at extremely low concentrations **Calculating the pH of weak Acids and Weak Bases** To calculate the pH of a solution of a weak acid or a weak base we must know the dissociation constant for each. Calculate the pH of a 0.1 M solution of ethanoic acid given that the value of K~a~ is 1.8 x 10^-5^ at 25^o^C. CH~3~COOH ↔ CH~3~COO^-^ + H^+\ ^Initially \[0.1\] \[0\] \[0\]\ At equil. \[0.1 -- x\] \[x\] \[x\] K~a~ = \[H^+^\]\[A^-^\]\ \[HA\] K~a~ = (x)(x)\ (0.1 -- x) Assume that 0.1- x = 0.1 (x)(x)\ 0.1 = 1.8 x 10^-5^ x^2^ = 1.8 x 10^-5^ x 0.1\ x = [\$\\sqrt{{1.8\\ x\\ 10}\^{- 5\\ }}x\\ 0.1\$]{.math.inline} = 1.3 x 10^-3^ pH = -log~10~\[H^+^\] = -log~10~(1.3 x 10^-3^) = 2.9 pH = 2.9 ^\ ^\[H^+^\] = [\$\\sqrt{Ka\\text{\\ x\\ M}\\text{acid}}\$]{.math.inline} where M~acid~ is the concentration of the acid in moles per litre \[OH^-^\] = [\$\\sqrt{Kb\\text{\\ x\\ M}\\text{base}}\$]{.math.inline} where M~base~ is the concentration of the base in moles per litre Calculate the pH of a 0.1 M solution of methanoic acid given that K~a~ for this acid is 2.1 x 10^-4^ \[H^+^\] = [\$\\sqrt{Ka\\text{\\ x\\ M}\\text{acid}}\$]{.math.inline}\ [\$\\sqrt{2.1\\ x\\ 10 - 4\\ x\\ 0.1}\$]{.math.inline}\ [\$\\sqrt{0.000021}\$]{.math.inline}\ [\$\\sqrt{0.0046}\$]{.math.inline}\ pH = -log~10~\[H^+^\] = -log~10~(0.0046) = 2.34 pH = 2.34 **Acid-Base Indicators** \*\*Definition: An acid-base indicator is a substance that changes colour according to the pH of the solution in which it is placed\*\* [Indicator as a weak acid:]\ HIn ⇌ H^+^ + In^-^\ Colour 1 Colour 2 Le Chatelier's Principle\ - Adding H^+^ (acid) will favour the reverse reaction, Colour 1 will be seen.\ - Adding OH^-^ (base) will react with the H^+^ to form water. This removes H^+^ from the system, the forward reaction will be favoured, Colour 2 will be seen. [Indicator as a weak base:]\ XOH ⇌ X^+^ + OH^-^\ Colour 1 Colour 2 Le Chatelier's Principle\ - Adding OH^-^ (base) will favour the reverse reaction, Colour 1 will be seen.\ - Adding H^+^ (acid) will react with the OH^-^ to form water. As this removes OH^-^ from the system, the forward reaction will be favoured, colour 2 will be seen. \*\*Definition: The range of an indicator is the pH interval over which there is a clear change of colour for that indicator.\*\* Indicators are chosen based on the pH range in which the colour changes, for example: Name of Indicator Approx. Range Acid Colour (Lower pH) Base Colour (Higher pH) ------------------- --------------- ------------------------ ------------------------- Methyl Orange 3 -- 5 Red Yellow Litmus 5 -- 8 Red Blue Phenolphthalein 8 - 10 colourless pink **[pH Titration]** A pH titration is an experiment to follow the pH changes as titration is carried out. \- The base is placed in the burette and the acid is placed in the beaker\ - The acid is kept stirred by placing a magnetic stir bar in the beaker\ - A pH sensor is placed in the acid solution. This measures the pH as the base is added from the burette to the acid in the beaker\ - The pH sensor is connected to a computer and the pH readings are recorded by software in the computer\ - A graph of pH versus time is plotted for the solution in the beaker. The graph is called a pH titration curve [Titrating a Strong Acid against a Strong Base] \- 0.1 M NaOH is slowly added to 25 cm^3^ of 0.1 M HCl.\ - Litmus, methyl orange and phenolphthalein can be used as indicators\ - There is a large jump of pH at the end point, this is because of the addition of one drop of base at the end point\ (2) [Titrating a Strong Acid against a Weak Base] \- 0.1 M NH~3~ solution is slowly added to 25 cm^3^ of 0.01 M HCl\ - Methyl Orange is a suitable indicator as its [Titrating a Weak Acid against a Strong Base] \- 0.1 M NaOH I slowly added to 25 cm^3^ of 0.1 M CH~3~COOH\ - Curve is almost vertical from 7 -- 10 so we phenolphthalein is a suitable indicator [Titrating a Weak Acid against a Weak Base] \- 0.01 M NH~3~ is slowly added to 25 cm^3^ of 0.1 M CH~3~COOH\ - pH rises so gradually that there is no sudden jump in the pH value at any stage\ - not possible to detect the endpoint using an indicator **[Exam Questions]** [2012 -- HL -- Section B -- Question 10] \(b) Define an acid in terms of the Brønsted-Lowry theory. An acid is a proton donor\ What is a conjugate pair? Acid and base that differ by one proton.\ A certain water soluble acid-base indicator represented by HIn is a weak acid which dissociates as follows in water.\ HIn H^+^ + In^--^\ State and explain the colour observed when a few drops of a solution of the indicator are added to a 0.5 M NaOH solution. Purple -Hydroxyl ions remove hydrogen ions causing the reaction to shift forward\ Calculate the pH of 0.5 M NaOH solution\ (i) pOH = --log~10~\[OH^--^\]\ log~10~0.5 = 0.3\ 14 -- 0.3 = 13.7\ pH = 13.7\ (ii) a 0.1 M solution of the indicator, given that its K~a~ value is 2.0 × 10-5\ \[H^+^\] = [\$\\sqrt{Ka\\text{\\ x\\ M}\\text{acid}}\$]{.math.inline}\ [\$\\sqrt{2.0}\\text{x\\ }10\^{- 5}\$]{.math.inline} x 0.1 = 0.00141\ pH = --log~10~\[H^+^\] = --log~10~(0.00141) =\ pH = 2.85 [2011 -- HL -- Section B -- Question 7] 7\. Sulfuric acid is a strong dibasic acid. The formula HA represents a weak monobasic acid.\ (a) How do strong acids differ from weak acids in their behaviour in water according to\ (i) the Arrhenius theory,\ Strong: Almost completely dissociated to give hydrogen ions (H^+^) in solution. Weak: Only slightly dissociated to give hydrogen ions (H^+^) in solution\ (ii) the Brønsted-Lowry theory?\ Strong: Good proton donor.\ Weak: Poor proton donor\ (b) What is the conjugate base of\ (i) sulfuric acid, Sulfuric acid: HSO~4~ ^--^\ (ii) the weak acid HA? A^-^\ Which of these conjugate bases is the stronger? Explain. A^--^ -- conjugate base of weak acid.\ (c) Explain, by giving a balanced equation for its dissociation in water, that the conjugate base of sulfuric acid is itself an acid.\ HSO~4~ ^--^ + H~2~O → SO~4~ ^2--^ + H~3~O^+^\ (d) Define pH. pH = --log~10~ \[H^+^\] **[References]** 1. Bbc.com 2. Chemistrystudent.com