Calculate the pH of a solution obtained by mixing 127.53 mL of 0.16 M CH3COOH with 10.13 mL of 0.1 M NaOH.

Answer 1

CH3COOH + NaOH ==> CH3COONa + H2O .. balanced equation

moles CH3COOH present = 127.53 ml x 1 L / 1000 ml x 0.16 mol/L = 0.0204 moles

moles NaOH added = 10.13 ml x 1 L / 1000 ml x 0.1 mol / L = 0.00101 moles moles

moles of CH3COONa produced = 0.00101 moles

moles CH3COOH left over = 0.0204 - 0.00101 = 0.0194

This makes a buffer because now you have a weak acid (CH3COOH) and the conjugate base (CH3COO-)

The final volume is 127.53 + 10.13 = 137.66 mls = 0.1377 L

[CH3COOH] = 0.0194 mol / 0.1337 L = 0.145 M

[CH3COO-] = 0.00101 mol / 0.1337 L = 0.00755 M (rather low for a good buffer)

Use the Henderson Hasselbalch equation:

pH = pKa + log [CH3COO-] / [CH3COOH]

pH = 4.75 + log (0.00755 / 0.145) = 4.75 -1.28

pH = 3.47

(be sure to check all of the math, as I tried to do this in a hurry).

Also, mention to whoever gave you this problem that to use 5 significant figures in the volume of CH3COOH is useless when the concentration is give to only 2 sig. figs. Laughable, actually.