Question

acid buffer described in Problem 21, you discover that your m ion? laboratory is out of sodium acetate, but you do have sodium hydroxide. How much (in moles and grams) acetic acid and til the pH on of the sodium hydroxide do you need to make the buffer? f th ola . Another alternative. Your friend from another labora- ory was out of acetic acid, so tries to prepare the buffer in Problem 21 by dissolving 41.02 g of sodium acetate in water, carefully adding 180.0 ml of 1 M HCl, and adding more water to reach a total volume of 2 liters. What is the total concentration of acetate plus acetic acid in the solu- tion? Will this solution have pH 5.0? Will it be identical e diluted solution H 5.0 is on? with the desired buffer? If not, how will it differ?

number 23

Answer 1

Number of moles of sodium acetate - mass/mol.wt. -41.02 g/(82.0343 g/mol) - 0.500035 mol Number of moles of HCl = M*V= 1 M * 180 mL= 1 M * 0.180 L = 0.180 mol Sodium acetate reacts with HCl to produce acetic acid. CH3COOHCl CH3COOH C 1 0.500035 C -0.180 E 0.320035 0.180 -0.180 0 0 +0.180 0.180 Number of moles of acetate ions present in the solution 0.320035 mol Number of moles of acetic acid formed 0.180 mol Total volume of the solution = 2 L Conc. of acetate-moles/total volume= 0.320035 mol/2 L= 0.160017 M Conc. of acetic acid 0.180 mol/2 L 0.09 M Total concentration of the solution 0.1600.09 0.250 M pH of the solution is [base] pH = pKa+log acid pH-log Ka +log [CH,COo] [CH COOH] 0.160017 pH-log (1.8x10)+log pH-4.745 +0.249925 pH 4.995 pH of the solution -5.0 0.09