Question

What is the pH of a solution that is 0.64M H2SO4 and 1.7M HCOOH (formic acid)?

Express your answer using two decimal places.

Answer 1

1) Calculate [H₃O⁺] from H₂SO₄

k_a1 = large

k_a2 = 1.2 x 10^-2

k_a1 >> k_a2

k_a1 is large, so the sulfuric acid will completely disassociate into HSO₄^- and H₃O⁺ ions as a strong acid. It means that the concentration of HSO₄^- and H₃O⁺ are the same as that of H₂SO₄, which is 0.64M.

HSO4-(aq)

+

H2O(liq)

?

SO42-(aq)

+

H3O+(aq)

	HSO4-(aq)	SO42-(aq)	H3O+(aq)
Initial	0.64M	0M	0.64M
Change	-X M	+X M	+X M
Equilibrium	(0.64 - X) M	X M	(0.64 + x) M

Assuming X in the denominator is negligible. Then, solving for X

X = 0.0117 M

From the ICE table

[H₃O⁺] = (0.64 + x) M = (0.64 + 0.0117) M = 0.6517 M from H₂SO₄

2) Calculate [H₃O⁺] from HCOOH

HCOOH(aq)

+

H2O(liq)

?

HCOO-(aq)

+

H3O+(sq)

	HCOOH(aq)	HCOO-(aq)	H3O+(sq)
Initial	1.7M	0M	0M
Change	-X M	+X M	+X M
Equilibrium	(1.7 - X) M	X M	X M

Assuming X in the denominator is negligible. Then, solving for X

X = 0.0175 M

[H₃O⁺] = x M = 0.0175 M from HCOOH

3) Calculate the total [H₃O⁺]

[H₃O⁺]_total = (0.0175 + 0.6517) M = 0.6692 M

4) Calculate pH

pH = - log [H₃O⁺] = - log (0.6692) = 0.17