Question

Please help, PH 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

we have:

Molarity of HF = 0.683 M

Volume of HF = 158 mL

Molarity of NaOH = 0.244 M

Volume of NaOH = 158 mL

mol of HF = Molarity of HF * Volume of HF

mol of HF = 0.683 M * 158 mL = 107.914 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.244 M * 158 mL = 38.552 mmol

We have:

mol of HF = 107.914 mmol

mol of NaOH = 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

we have below equation to be used:

This is Henderson–Hasselbalch equation

pH = pKa + log {[conjugate base]/[acid]}

= 3.143+ log {0.122/0.2195}

= 2.89

Answer:  2.89