Question

A. What is the pH of a 1.0 L buffer made with 0.15 M formic acid (HCOOH, Ka = 1.8 x 10-4) and 0.25 M potassium formate, KHCOO?

B. What is the pH of the buffer solution in question #13 after 0.096 moles of HCl are added to it?

Answer 1

A)

we know that

for buffers

pH = pKa + log [conjugate base / acid]

also

pKa = -log Ka

so

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

pH = -log 1.8 x 10-4 + log [ 0.25 / 0.15]

pH = 3.966

so

pH of the buffer is 3.966

B)

now

we know that

moles = concentration x volume (L)

so initially

moles of HCOOH = 0.15 x 1 = 0.15

moles of HCOO- = 0.25 x 1 = 0.25

now

moles of HCl added = 0.096

now

the reaction is

HCOO- + H+ --> HCOOH

we can see that

moles of HCOO- reacted = moles of HCl added = 0.096

moles of HCOOH formed = moles of HCl added = 0.096

now finally

moles of HCOO- = 0.25 - 0.096 = 0.154

moles of HCOOH = 0.15 + 0.096 = 0.246

now

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

pH = -log 1.8 x 10-4 + log [ 0.154 / 0.246]

pH = 3.5413

so

pH after addition of HCl is 3.5413