Calculate the pH during the titration of 20.00 mL of 0.1000 M HCOOH(aq) with 0.2000 M NaOH(aq) after 5.5 mL of the base have been added.

Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M HCOOH(aq) with 0.2000 M NaOH(aq) after 5.5 mL of the base have been added.

K_a of formic acid = 1.8 x 10^-4.

the answer is 3.83

please show your steps, thank you :)

I kind of understand this type of question but I don't see where Ka comes in? Please show how to complete this problem, thank you :)

Answer 1

moles HCOOH initially present = 0.02000 L x 0.1000 mol/L = 0.002000 moles

moles NaOH added = 0.0055 L x 0.2000 mol/L = 0.0011 moles

HCOOH + NaOH ===> HCOONa + H₂O

0.002mol....0.0011mol.....0............0........Initial

-0.0011......-0.0011........+0.0011............Change

0.0009.........0..............0.0011...............Equilibrium

Final volume = 25.5 ml = 0.0255 L

Final [HCOOH] = 0.0009 mol/0.0255 L = 0.03529 M

Final [HCOONa] = 0.0011 mol/0.0255 L = 0.04314 M

Since you have formed a buffer solution which now contains the weak acid (HCOOH) and the conjugate base of that acid (HCOONa), you can use the Henderson Hasselbalch equation to find the pH. But first we should find the pKa from the Ka. pKa = -log Ka = - log 1.8x10^-4 = 3.745

pH = pKa + log [conjugate base]/[acid] = 3.745 + log [0.04314]/[0.03529]

pH = 3.745 + 0.0872

pH = 3.83