Question

1) A 35.4 mL sample of a 0.394 M aqueous nitrous acid solution is titrated with a 0.222 M aqueous solution of potassium hydroxide. How many milliliters of potassium hydroxide must be added to reach a pH of 3.167?

2) Calculate the pH of a 0.1510 M aqueous solution of potassium dihydrogen phosphate, KH₂PO₄.

Answer 1

pH =3.16

[H+] = 6.91×10^-4M

No of mole in 35.4 ml = 2.4 × 10^-5

HNO2. +. KOH ---------> KNO2 + H2O

1 mole of KOH react with 1 mole of KOH

Volume of acid = 35.4 ml

Molarity of acid = 0.394M

No of mole =( 0.394mole/1000ml)× 35.4ml = 0.01394 mole

No of mole of H+ = 0.01394

No of mole of H+ reacted = 0.01394 - 0.000024= 0.013916

no of mole KOH needed = 0.013916

Molarity of KOH = 0.222M

Volume of KOH needed = 62.68 ml

2) Ka for KH2PO4 = 6.3 × 10^-8

H2PO4- < ---------> [HPO4^2-][H+]/[H2PO4-]

Ka = [HPO4^2-][[H+]/[ H2PO4-]

6.3 × 10^-8 = X^2/0.151

X = 9.75 × 10^-5

[H+] = 9.75 ×10^-5

pH = 4.01