Question

A 21.8 mL sample of 0.337 M ammonia, NH3, is titrated with 0.291 M hydroiodic acid. After adding 35.3 mL of hydroiodic acid, the pH is .

Answer 1

Given:

M(HI) = 0.291 M

V(HI) = 35.3 mL

M(NH3) = 0.337 M

V(NH3) = 21.8 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.291 M * 35.3 mL = 10.2723 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.337 M * 21.8 mL = 7.3466 mmol

We have:

mol(HI) = 10.2723 mmol

mol(NH3) = 7.3466 mmol

7.3466 mmol of both will react

excess HI remaining = 2.9257 mmol

Volume of Solution = 35.3 + 21.8 = 57.1 mL

[H+] = 2.9257 mmol/57.1 mL = 0.0512 M

use:

pH = -log [H+]

= -log (5.124*10^-2)

= 1.2904

Answer: 1.29