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Consider the titration of 30.0 mL of 0.283 M weak base B (Kb = 1.3 x...

Consider the titration of 30.0 mL of 0.283 M weak base B (Kb = 1.3 x 10⁻¹⁰) with 0.150 M HI. What would be the pH of the solution after the addition of 100.0 mL of HI?

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Answer #1

Given:
M(HI) = 0.15 M
V(HI) = 100 mL
M(B) = 0.283 M
V(B) = 30 mL


mol(HI) = M(HI) * V(HI)
mol(HI) = 0.15 M * 100 mL = 15 mmol

mol(B) = M(B) * V(B)
mol(B) = 0.283 M * 30 mL = 8.49 mmol



We have:
mol(HI) = 15 mmol
mol(B) = 8.49 mmol

8.49 mmol of both will react
excess HI remaining = 6.51 mmol
Volume of Solution = 100 + 30 = 130 mL
[H+] = 6.51 mmol/130 mL = 0.0501 M

use:
pH = -log [H+]
= -log (5.008*10^-2)
= 1.3004
Answer: 1.30

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