Question #cb24c

1 Answer
Stefan V.
Oct 24, 2015

pH=2.54

Explanation:

Your strategy here is to use an ICE table to find the equilibrium concentration of the hydronium ions, H3O+, formed in solution by the partial ionization of formic acid (methanoic acid), HCO2H, a weak acid.

To get the acid dissociation constant, Ka, use the given pKa

Ka=10−pKa

Ka=10−3.75=0.00017778=1.78⋅10−4

So, use an ICE table to get the equilibrium concentration of the hydronium ions

CHCO2H(aq]+H2O(l]⇌CHCO−2(aq] + H3O+(aq]

I 0.05 0 0
C (−x) (+x) (+x)
E 0.05−x x x

By definition, the acid dissociation constant will be

Ka=[H3O+]⋅[CHO−2][CHO2H]

Ka=x⋅x0.05−x=1.78⋅10−4

Since the initial concentration of the acid is relatively small, you cannot use the approximation

0.05−x≈0.05

This means that will have to solve for x by using a quadratic equation

x2=1.78⋅10−4⋅(0.05−x)

x2=8.9⋅10−6−1.78⋅10−4x

x2+1.78⋅10−4x−8.9⋅10−6=0

This quadratic equation will produce two solutions, one positive and one negative. Since x symbolizes concentration, the only solution that will have chemical and physical significance will be the positive one

x=0.0028956

This means that the concentration of hydronium ions will be

[H3O+]=x=0.0028956 M

The pH of the solution will be

pH=−log([H3O+])

pH=−log(0.0028956)=2.54

SIDE NOTE You can try to solve by using the approximation

0.05−x≈0.05

the pH of the solution will be similar, but the approximation error will be greater than 5%, which indicates that the approximation is not justified.

0.05−0.002983=0.04702

The error is

|0.04702−0.05|0.05×100=5.96%

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