Question

500.0 mL of 0.130 M NaOH is added to 575 mL of 0.200 M weak acid (Ka = 7.35 × 10-5). What is the pH of the resulting buffer?

Answer 1

we have:

Molarity of HA = 0.2 M

Volume of HA = 575 mL

Molarity of NaOH = 0.13 M

Volume of NaOH = 500 mL

mol of HA = Molarity of HA * Volume of HA

mol of HA = 0.2 M * 575 mL = 115 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.13 M * 500 mL = 65 mmol

We have:

mol of HA = 115 mmol

mol of NaOH = 65 mmol

65 mmol of both will react

excess HA remaining = 50 mmol

Volume of Solution = 575 + 500 = 1075 mL

[HA] = 50 mmol/1075 mL = 0.0465M

[A-] = 65/1075 = 0.0605M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 7.35*10^-5

pKa = - log (Ka)

= - log(7.35*10^-5)

= 4.134

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 4.134+ log {6.047*10^-2/4.651*10^-2}

= 4.25

Answer: 4.25