Answer to Question #136081 in General Chemistry for Lystra

Question

What is the ph of a 1.15M solution of methanoic acid? Ka= 1.8*10^4

Answer 1

The chemical equation:

_________HCOOH + H₂O <=> HCOO^- + H₃O⁺

ICE table: 1.15________________0______0

___________-x________________ +x_____ +x

__________1.15-x______________ x ______x

$Ka = \frac{[HCOO-][H3O+]}{[HCOOH]} = \frac{x^2}{1.15 – x} = 1.8\times10^-4$

Check for negligibility:

$\frac{1.15}{1.8*10^{-4}} = 6388 > 400$

x<<1.15

Then

$\frac{x^2}{1.15} = 1.8\times10^{-4}$

$x^2 = 1.15\times1.8\times10^{-4}$

x = 0.0143

[H₃O⁺] = 0.0143 mol/L

pH = -log[H₃O⁺] = -log0.0143 = 1.84

Answer: 1.84