Question

How many grams of solid NaOH (Mol. Wt. = 40 g mol-1) must be added to a 0.15 L solution of 0.02 mol L–1 glycine hydrochloride to adjust the pH to 2.65? (pKa1 = 2.35, pKa2 = 9.78)

(A) 79.92 g

(B) 7.992 g

(C) 0.799 g

(D) 0.079 g

(E) 0.008 g

Answer 1

Answer is (D) 0.079g

Formula used to solve this problem is,

pH = pKa + log([A-]/[HA])

Reaction is,

H₂N-CH₂-COOH =  H₂N-CH₂-COO^- + H₃N-CH₂-COO⁺

^HA   A^- (charged) neutral

before reaction 0.02 0 0

after reaction 0.02-x x x

Now,

pH = pKa + log([A-]/[HA])

2.65 = 2.35 + log (x / (0.02-x))

0.3 =  log (x / (0.02-x))

(x / (0.02-x)) = 1.995

x = 0.0133 mole L^-1

Amount of NaOH in 0.0133 mole = 40 * 0.0133 = 0.5324 g

Amount of NaOH in 0.15 L solution = 0.5324 * 0.15 = 0.0798 g ~ 0.079 g

Thus, 0.079 g of NaOH is required to adjust the pH at 2.65.