Question

A student is asked to determine the value of Ka for hydrocyanic acid by titration with barium hydroxide. The student begins titrating a 40.8 mL sample of a 0.552 M aqueous solution of hydrocyanic acid with a 0.317 M aqueous barium hydroxide solution, but runs out of standardized base before reaching the equivalence point. The student's last observation is that when 22.4 milliliters of barium hydroxide have been added, the pH is 9.672. What is Ka for hydrocyanic acid based on the student's data?

Answer 1

Ka = hydrocyanic acid

2HCN + Ba(OH)2 --> 2H2O + Ba(CN)2

V = 40.8 mL of acid

M = 0.552 M of acid

M = 0.317 M of base

no complete!

so

V = 22.4 mL of Ba(OH)2 are used, then pH = 9.672

note that this is a weak acid, addition of Ba(OH)2 will form a buffer

so

pH = pKA + log(CN- /HCN)

we need to find CN- and HCN in solution

9.672 = pKA + log(CN- /HCN)

mmol initially of HCN = MV = 40.8*0.552 = 22.5216 mmol of HCN

mmol of base added = MV = 0.317*22.4 = 7.1008 mmol

mmol of OH- = 2*mmol ov base = 7.1008*2 = 14.2016 mmol

so...

there will be reaction and conjugate formation

mmol of acid left = 22.5216 -14.2016 = 8.32 mmol

mmol of conjguate formed = 14.2016 mmol

now... substitute all in equation

9.672 = pKA + log(CN- /HCN)

9.672 = pKA + log(14.2016 /8.32 )

pKa = 9.672 -  log(14.2016 /8.32 )

pKa =9.439 HCN

Ka = 10^-pKA = 10^-9.439

Ka = 3.639*10^-10