Question

500.0 mL of 0.100 M NaOH is added to 575 mL of 0.250 M weak acid (Ka = 3.61 × 10-5). What is the pH of the resulting buffer?

Answer 1

we have:

Molarity of HA = 0.250 M

Volume of HA = 575 mL

Molarity of NaOH = 0.1 M

Volume of NaOH = 500 mL

mol of HA = Molarity of HA * Volume of HA

mol of HA = 0.25 M * 575 mL = 143.75 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.1 M * 500 mL = 50 mmol

We have:

mol of HA = 143.75 mmol

mol of NaOH = 50 mmol

50 mmol of both will react

excess HA remaining = 93.75 mmol

Volume of Solution = 575 + 500 = 1075 mL

[HA] = 93.75 mmol/1075 mL = 0.0872M

[A-] = 50/1075 = 0.0465M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 3.61*10^-5

pKa = - log (Ka)

= - log(3.61*10^-5)

= 4.4425

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 4.4425+ log {0.0465/0.0872}

= 4.1695

Answer: 4.17