Question

1. What is the pH at the equivalence point in the titration of a 20.5 mL sample of a 0.437 M aqueous hydrocyanic acid solution with a 0.435 M aqueous potassium hydroxide solution?

pH =

2. When a 15.4 mL sample of a 0.408 M aqueous nitrous acid solution is titrated with a 0.322 M aqueous sodium hydroxidesolution, what is the pH at the midpoint in the titration?

pH =

Answer 1

1)

HCN + KOH ----------> KCN + H2O

initially millimoles of HCN = 20.5 x 0.437 = 8.96

8.96 millimoles of KOH must be added to reach equivalence point .

8.96 = V x 0.435

V = 20.60 mL KOH must be added

total volume = 20.5 + 20.6 = 41.1 mL

[KCN] = 8.96 / 41.1 = 0.218 M

KCN is salt of weak acid for such salts

pH = 1/2 [pKw + pKa + log C]

pKa of HCN = 9.2 standard value

pH = 1/2 [14 + 9.2 + log 0.218]

pH = 1/2 [22.54]

pH = 11.27

pH = 11.3

2) HNO2 + NaOH ------------> NaNO2 + H2O

titration mid point means 50 % of reaction over

means resulting dolution contain 50% HNO2 and 50 % NaNO2

means [HNO2] = [NaNO2]

in such case for any titration

pH = pKa

pKa of HNO2 = 3.15

pH = 3.15