BUFFER QUESTION CAN SOME ON HELP?

Question

1 QUESTIOIN///

A FORMIC ACID BUFFER CONTAINIG 0.50 M HCOOH ABD 0.50 M HCOONa IS PREPARED BEFORE TITRATION(Ka HCOOH=1.8X10^-4) WHAT IS THE pH AFTER THE ADDITION OF 10.0 mL OF 0.100 M NaOH TO THE 100.0mL BUFFER?

Answer 1

Moles HCOOH = 0.50 M x 0.100 L = 0.050

Moles HCOO- = 0.50 M x 0.100 L = 0.050

Moles NaOH = 0.100 M x 0.0100 L =0.00100

The effect of the added OH- is to concert HCOOH to HCOO- via the net reaction

OH- + HCOOH >> HCOO- + H2O

moles HCOOH = 0.050 - 0.00100 =0.049

moles HCOO- = 0.050 + 0.00100 = 0.051

total volume = 100 + 10 = 110 mL = 0.110 L

Concentration HCOOH = 0.049 / 0.110 = 0.455 M

concentration HCOO- = 0.051 / 0.110 = 0.464 M

pKa = - log Ka = 3.74

pH = pKa + log [HCOO-] / [HCOOH] =

= 3.74 + log 0.464 / 0.455 = 3.75