Question

An analytical chemist is titrating 75.6mL of a 0.8200M solution of isopropylamine CH32CHNH2 with a 0.5300M solution of HIO3. The pKb of isopropylamine is

3.33. Calculate the pH of the base solution after the chemist has added 132.8mL of the HIO3 solution to it.

Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HIO3 solution added.

Round your answer to 2 decimal places.

An analytical chemist is titrating 75.6 mL of a 0.8200 M solution of isopropylamine ((CH3), CHNH2 with a 0.5300 M solution of HIO3. The pK, of isopropylamine is 3.33. Calculate the pH of the base solution after the chemist has added 132.8 mL of the HIO, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HIO, solution added. Round your answer to 2 decimal places. pH = 1 x 6 ?

Answer 1

Given:

M(HIO3) = 0.53 M

V(HIO3) = 132.8 mL

M((CH3)2CHNH2) = 0.82 M

V((CH3)2CHNH2) = 75.6 mL

mol(HIO3) = M(HIO3) * V(HIO3)

mol(HIO3) = 0.53 M * 132.8 mL = 70.384 mmol

mol((CH3)2CHNH2) = M((CH3)2CHNH2) * V((CH3)2CHNH2)

mol((CH3)2CHNH2) = 0.82 M * 75.6 mL = 61.992 mmol

We have:

mol(HIO3) = 70.384 mmol

mol((CH3)2CHNH2) = 61.992 mmol

61.992 mmol of both will react

excess HIO3 remaining = 8.392 mmol

Volume of Solution = 132.8 + 75.6 = 208.4 mL

[H+] = 8.392 mmol/208.4 mL = 0.0403 M

use:

pH = -log [H+]

= -log (4.027*10^-2)

= 1.395

Answer: 1.40