Medical Chemistry Lecture 2 PDF

Al-Mustafa University College Department of dentistry 1st Class Medical Chemistry Dr. Saeed Al-Hamash © 2012 Pearson Education, Inc. Concentration of Solution Solvent Solute The concentration of a solution is a measure of the amount of solute that is dissolved in a g...

Al-Mustafa University College Department of dentistry 1st Class Medical Chemistry Dr. Saeed Al-Hamash © 2012 Pearson Education, Inc. Concentration of Solution Solvent Solute The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. –A dilute solution is one that contains a small amount of solute. –A concentrated solution contains a large amount of solute. What we need is a way of quantifying the concentration of a solution! Molarity(M) is the number of moles of solute dissolved in one liter of solution. To calculate the molarity of a solution, divide the moles of solute by the volume of the solution.  Concentration of Solution Moles of solute Mol Molarity (M) = = Liter of solution L Parts ratio= amount of solute (g or ml) (10 ) or (10 ) or (10 ) 2 6 9 amount of solution (g or ml) Mole Fraction()= Moles of solute Total moles of solution Molality (m) = Moles of solute Kilograms of solvent  Molarity NaCl Molarity Example Problem 1 12.6 g of NaCl are dissolved in water making 344mL of solution. Calculate the molar concentration.  1molNaCl  12.6 g NaCl   moles solute  58.44 gNaCl  M= = L solution  1L  344 mL   solution  1000mL  = 0.627 M NaCl  Molarity NaCl Molarity Example Problem 2 How many moles of NaCl are contained in 250.mL of solution with a concentration of 1.25 M? moles solute M= L solution  1L  250. mL   = 0.250 L solution  1000mL  Volume x concentration = moles solute (0.25 ) x 1.25= 0.3125 moles  Molarity NaCl Molarity Example Problem 3 What volume of solution will contain 15 g of NaCl if the solution concentration is 0.75 M? moles solute therefore the M= solution contains L solution  1 mol NaCl  0.75 mol NaCl 15 g NaCl   = 0.257 mol  58.44 g NaCl  1 L solution moles solute ÷ concentration = volume solution  1 L solution  0.257 mol NaCl   = 0.34 L solution  0.75 mol NaCl  Example : If 5 liters of water is added to two moles of glucose to make a solution, the concentration (molarity) is said to be 0.4 M Sol : 2 mol of glucose / 5 liters of solution =0.4 M Example : What is the molarityof 600. mLof potassium iodide solution that contains 5.50 moles of the solute? Sol : M = mol solute / L sol. M = 5.50 mol /.600 L M = 9.16 molar H.W: Example : A saline solution contains 0.90 g of NaCl per 100 mL of solution. What is its molarity? Sol. : 1) convert grams to moles... 0.90 g NaCl/ 58.5 g/mol = 0.015 mol NaCl 2) convert mLto L and solve... 0.015 mol /.100 L = 0.15 M Example : What volume of a 4.0 M solution would contain 15.5 moles of sodium thiosulfate? Sol : M = mol solute / L solution... so... L = mol / M L = 15.5 mol / 4.0 M L = 3.9 liters of solution % Concentration mass solute  % (w/w) = x 100 mass solution  % (w/v) = mass solute x 100 volume solution  % (v/v) = volume solute x 100 volume solution Mass and volume units must match. (g & mL) or (Kg & L)  % Concentration Example Problem 1 What is the concentration in %w/v of a solution containing 39.2 g of potassium nitrate in 177 mL of solution? mass solute 39.2 g % (w/v) =  100  100 = 22.1 % w/v volume solution 177 mL Example Problem 2 What is the concentration in %v/v of a solution containing 3.2 L of ethanol in 6.5 L of solution? volume solute 3.2 L % (v/v) =  100  100 = 49 % v/v volume solution 6.5L  % Concentration Example Problem 3 What volume of 1.85 %w/v solution is needed to provide 5.7 g of solute? 1.85 g solute % (w/v) = 100 mL solution We know: We want to get: g solute g solute and mL solution mL solution  100 mL solution  5.7 g solute   = 310 mL Solution  1.85 g solute  g solute ÷ concentration = volume solution  Parts per million/billion (ppm & ppb) mass solute mg  ppm = × 106 or = ppm volume solution L  ppb = mass solute g × 109 or = ppb volume solution L AND Mass and volume units must match. For very low (g & mL) or (Kg & L) concentrations: ng parts per trillion = ppt L  ppm & ppb Example Problem 1 An Olympic sized swimming pool contains 2,500,000 L of water. If 1 tsp of salt (NaCl) is dissolved in the pool, what is the concentration in ppm? 1 teaspoon = 6.75 g NaCl or g solute mg solute ppm = ×106 ppm = mL solution L solution 6.75 g 6 6.75 g  1000 1 g  mg ppm = ×10 ppm = 2.5×10 L  1000 6 1 L  mL 2.5×106 L ppm = 0.0027 ppm = 0.0027  ppm & ppb Example Problem 2 An Olympic sized swimming pool contains 2,500,000 L of water. If 1 tsp of salt (NaCl) is dissolved in the pool, what is the concentration in ppb? 1 teaspoon = 6.75 g NaCl or g solute  g solute ppb = ×109 ppb = mL solution L solution ppb = 6.75 g ×109 ppb = 6.75 g  106 mg 1 g  2.5×10 L  1000 6 1 L  mL 2.5×106 L ppb = 2.7 ppb = 2.7  Mole Fraction B A A B A A A B B A B A A A Mole Fraction () moles of A A  A = sum of moles of all components A + B moles of B B  B = sum of moles of all components A + B Since A + B make up the entire mixture, their mole fractions will add up to one.  A  B  1.00 Mole Fraction Example Problem 1 In our glass of iced tea, we have added 3 tbsp of sugar (C12H22O11). The volume of the tea (water) is 325 mL. What is the mole fraction of the sugar in the tea solution? (1 tbsp sugar ≈ 25 g) First, we find the moles of both the solute and the solvent.  1 mol C12 H 22 O11   1 mol H 2 O  75.g C12 H 22O11   = 0.219 mol 325mL H 2O   = 18.1 mol  342 g C12 H 22 O11   18.0 g H 2 O  Next, we substitute the moles of both into the mole fraction equation. χ sugar = moles solute total moles solution = 0.219 mol sugar (0.219 mol + 18.1 mol)  0.012 Mole Fraction Example Problem 2 Air is about 78% N2, 21% O2, and 0.90% Ar. What is the mole fraction of each gas? First, we find the moles of each gas. We assume 100. grams total and change each % into grams.  1 mol N2   1 mol O2  78g N2   = 2.79 mol 21g O2   = 0.656 mol  28 g N2   32 g O 2   1 mol Ar  Next, we substitute the moles of each 0.90g Ar   = 0.0225 mol  40. g Ar  into the mole fraction equation. χ = moles N 2 N2 total moles χ = moles O 2 O2 total moles χ Ar = moles Ar total moles 2.79 mol N 2 0.656 mol O2 0.0225 mol Ar = = = (2.79 + 0.656 + 0.0225) (2.79 + 0.656 + 0.0225) (2.79 + 0.656 + 0.0225)  0.804  0.189  0.00649  Molal (m) Example Problem 1 If the cooling system in your car has a capacity of 14 qts, and you want the coolant to be protected from freezing down to -25°F, the label says to combine 6 quarts of antifreeze with 8 quarts of water. What is the molal concentration of the antifreeze in the mixture? mol solute antifreeze is ethylene glycol C2H6O2 m= 1 qt antifreeze = 1053 grams Kg solvent 1 qt water = 946 grams  1053 g C2 H 6O2   1mol C2 H 6O 2  6 Qts      1 Qt C H O 2 6 2   62.1 g C 2 H 6O 2  m= = 13 m  946 g H 2O   1 Kg  8 Qts    1000 g   1 Qt H 2 O   