Question

500.0 mL of 0.110 M NaOH is added to 585 mL of 0.200 M weak acid (Ka = 7.30 × 10-5). What is the pH of the resulting buffer?

Answer 1

we have:

Molarity of HA = 0.2 M

Volume of HA = 585 mL

Molarity of NaOH = 0.11 M

Volume of NaOH = 500 mL

mol of HA = Molarity of HA * Volume of HA

mol of HA = 0.2 M * 585 mL = 117 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.11 M * 500 mL = 55 mmol

We have:

mol of HA = 117 mmol

mol of NaOH = 55 mmol

55 mmol of both will react

excess HA remaining = 62 mmol

Volume of Solution = 585 + 500 = 1085 mL

[HA] = 62 mmol/1085 mL = 0.0571M

[A-] = 55/1085 = 0.0507M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 7.3*10^-5

pKa = - log (Ka)

= - log(7.3*10^-5)

= 4.137

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 4.137+ log {5.069*10^-2/5.714*10^-2}

= 4.085

Answer: 4.085