Question

What is the pH of a solution that results from adding 158 mL of 0.244 M NaOH to 158 mL of 0.683 M HF? (Ka of HF = 7.2E-4) Ph= _________

Will give credit for the answer, thank you!

Answer 1

Given:
M(HF) = 0.683 M
V(HF) = 158 mL
M(NaOH) = 0.244 M
V(NaOH) = 158 mL


mol(HF) = M(HF) * V(HF)
mol(HF) = 0.683 M * 158 mL = 107.914 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.244 M * 158 mL = 38.552 mmol

38.552 mmol of both will react

excess HF remaining = 69.362 mmol
Volume of Solution = 158 + 158 = 316 mL
[HF] = 69.362 mmol/316 mL = 0.2195M

[F-] = 38.552/316 = 0.122M

They form acidic buffer
acid is HF
conjugate base is F-


Ka = 7.2*10^-4

pKa = - log (Ka)
= - log(7.2*10^-4)
= 3.143

use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.143+ log {0.122/0.2195}
= 2.89


Answer: 2.89