What is the pH of a 6.2⋅10−3 M solution of HF?

1 Answer
Dr. Hayek
Jun 5, 2016

pH=2.77

Explanation:

The acid dissociation reaction for HF is the following:

HF→H++F−

The Ka expression is: Ka=[H+][F−][HF]=6.6×10−4

The pH of this solution could be found by: pH=−log[H+].

We will need to find the [H+] using ICE table:

HF → H++ F−
Initial 6.2×10−3M 0M 0M
Change −xM +xM +xM
Equilibrium(6.2×10−3−x)M xM xM

now, we can replace the concentrations by their values in the expression of Ka:

Ka=[H+][F−][HF]=6.6×10−4=x×x6.2×10−3−x

Solve for x=1.7×10−3M

⇒pH=−log[H+]=−log(1.7×10−3)=2.77

Acids & Bases | pH of a Weak Acid.

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