Question

a) We place 0.852 mol of a weak acid, HA, and 12.6 g of NaOH in enough water to produce 1.00 L of solution. The final pH of this solution is 4.04 . Calculate the ionization constant, K_b, of A^-(aq).

b) HA is a weak acid with a pK_a value of 3.73 . What is the pH of a buffer solution that is 0.395 M in HA and 0.833 M in NaA? What is its pH after the addition of 12.0 g of HCl to 2.000 L of this buffer solution? Assume the volume remains fixed at 2.000 L.

Answer 1

(a) Molar mass of NaOH = 40 g/mol. So, 12.6 g NaOH = (12.6 / 40) mol NaOH = 0.315 mol NaOH. So, after adding base, moles of salt = 0.315 mol and that of acid = (0.852 - 0.315) mol = 0.537 mol. Since the solution has a volume 1.00 L, thus [NaA] = 0.315 M and [HA] = 0.537 M. From Henderson-Hasselbalch equation we know,

So, pK_b of A^- = 14 - 4.27 = 9.73. Thus, K_b = 10 ^{- 9.73} = 1.862 x 10^-10

So, the answer is 1.862 x 10^-10.

(b) Again from HH equation we have,

Thus,

Now, if we add 12.0 g HCl, which has a molar mass = 36.45 g/mol, then moles of HCl added = (12.0 / 36.5) mol = 0.329 mol HCl.

Now, amount of NaA

2.000 L 0.833 M = (2.000 x 0.833) LxM = 1.666 mol, due to addition of HCl this will be converted to (1.666 - 0.329) mol = 1.337 mol. So, [NaA] = 1.337 / 2.000 M = 0.6685 M.

Similarly amount of HA

2.000 L 0.395 M = (2.000 x 0.395) LxM = 0.790 mol, due to addition of HCl this will be converted to (0.790 + 0.329) mol = 1.119 mol. So, [HA] = 1.119 / 2.000 M = 0.5595 M.

So,