Question

Calculate the pH when 15.0 mL of 0.20 M Benzoic Acid (HC₇H₅O₂ K_a = 6.5x10^-5) is titrated with 20.0 mL of 0.15 M sodium hydroxide (NaOH)

Can you exaplain how to solve this question?

Answer 1

we have:

Molarity of C6H5COOH = 0.2 M

Volume of C6H5COOH = 20 mL

Molarity of kOH = 0.15 M

Volume of kOH = 20 mL

mol of C6H5COOH = Molarity of C6H5COOH * Volume of C6H5COOH

mol of C6H5COOH = 0.2 M * 20 mL = 4 mmol

mol of kOH = Molarity of kOH * Volume of kOH

mol of kOH = 0.15 M * 20 mL = 3 mmol

We have:

mol of C6H5COOH = 4 mmol

mol of kOH = 3 mmol

3 mmol of both will react

excess C6H5COOH remaining = 1 mmol

Volume of Solution = 20 + 20 = 40 mL

[C6H5COOH] = 1 mmol/40 mL = 0.025M

[C6H5COO-] = 3/40 = 0.075M

They form acidic buffer

acid is C6H5COOH

conjugate base is C6H5COO-

Ka = 6.3*10^-5

pKa = - log (Ka)

= - log(6.3*10^-5)

= 4.201

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 4.201+ log {7.5*10^-2/2.5*10^-2}

= 4.68

Answer: 4.68