Question

1.

a. What is the pH of a solution prepared by dissolving 54 mmol of a weak base in 350 mL of a 0.199-M solution of its conjugate acid if pKa = 5.57 for the acid.

b. How many mL (to the nearest mL) of 0.223-M KF solution should be added to 590. mL of 0.137-M HF to prepare a pH = 2.90 solution?

Answer 1

a. What is the pH of a solution prepared by dissolving 54 mmol of a weak base in 350 mL of a 0.199-M solution of its conjugate acid if pKa = 5.57 for the acid.

note that we are talking about a weak base, but the pKa is given.. assume this is correctly given (this is not typically, given...)

then

pH = pKa + log(A-/HA)

let B = weak base, BH+ = conjguate acid

pH = 5.57 + log(B/BH+)

[B] = 54/350 = 0.1543

[BH+] = 0.199

pH = 5.57 + log(0.1543/0.199)

pH = 5.4595

b. How many mL (to the nearest mL) of 0.223-M KF solution should be added to 590. mL of 0.137-M HF to prepare a pH = 2.90 solution?

pH = pKa + log(F-/HF)

mmol of HF = MV = 590*0.137 = 80.83

2.90 = 3.14 + log(mmol of F- / 80.83)

mmol of F- = 80.83 * 10^(2.90 - 3.14)

mmol of F- = 46.51

V of Kf = mmol/M = 46.51/0.223 = 208.56 mL of KF