Question

Calculate the [H⁺] of 25 mL of 0.025 M lactic acid (for which K_a = 1.37 × 10^–4) and 25 mL of 0.015 M NaOH. Please show all steps. Thanks.

Answer 1

Lets write lactic acid as HA

Given:

M(HA) = 0.025 M

V(HA) = 25 mL

M(NaOH) = 0.015 M

V(NaOH) = 25 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.025 M * 25 mL = 0.625 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.015 M * 25 mL = 0.375 mmol

We have:

mol(HA) = 0.625 mmol

mol(NaOH) = 0.375 mmol

0.375 mmol of both will react

excess HA remaining = 0.25 mmol

Volume of Solution = 25 + 25 = 50 mL

[HA] = 0.25 mmol/50 mL = 0.005M

[A-] = 0.375/50 = 0.0075M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 1.37*10^-4

pKa = - log (Ka)

= - log(1.37*10^-4)

= 3.8633

use:

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

= 3.8633+ log {0.0075/0.005}

= 4.0394

use:

pH = -log [H+]

4.039 = -log [H+]

[H+] = 9.13*10^-5 M

Answer: 9.13*10^-5 M