PLEASE HELP ME OUT ASAP !

Question

Calculate the pH of the solution when 0.50 g of NaOH is added to 200.0 mL of 0.100 M HCOOH (formic acid). it will be accepted there is no change in the volume of solution when solid substance is added (Ka=1.80x10-4) (Na:22.99 ; O:16.00 ; H:1.01 g/mol).

PLEASE WRITE THE SOLUTION WITH ALL STEPS

Answer 1

When you add NaOH to formic acid, you create a BUFFER.

HCOOH + NaOH ==> HCOONa + H2O

HCOOH + HCOONa = BUFFER (weak acid + salt of that acid)

To find the pH of such a buffer, we can use the Henderson Hasselbalch (HH) equation:

pH = pKa + log [HCOONa]/[HCOOH]

So we need to find the [HCOONa] and [HCOOH] and also find the pKa

pKa = -log Ka = -log 1.80x10^-4 = 3.74 = pKa

moles NaOH added = 0.50 g x 1 mol NaOH / 40 g = 0.0125 moles NaOH

moles HCOOH initially present = 200 ml x 1 L/1000 ml x 0.100 mol/L = 0.02 moles HCOOH

HCOOH + NaOH ===> HCOONa + H2O

0.02............0.0125..............0..................Initial

-0.0125......-0.0125.........+0.0125..........Change

0.0075.........0.................0.0125............Equillibrium

pH = pKa + [HCOONa]/[HCOOH]

pH = 3.74 + log (0.0625 M)/(0.0375) because the volume is 200 ml = 0.200 L

pH = 3.74 + 0.22

pH = 3.96

Could also have used:

pH = 3.74 + log (0.0125/0.0075) since dividing both moles by 0.2 L won't change the fraction.