The Ka of a weak acid is 1.48 x 10-5. Assuming the "x is small" approximation...

The Ka of a weak acid is 1.48 x 10-5. Assuming the "x is small" approximation is valid, what is the predicted pH of a 0.05 M solution of the weak acid? Provide your response to two decimal places.

Expert Solution

HA ------------> H^+ (aq) + A^- (aq)

I 0.05 0 0

C -x +x +x

E 0.05-x +x +x

Ka = [H+][A-]/[HA]

1.48*10^-5 = x*x/(0.05-x)

1.48*10^-5 *(0.05-x) = x^2

x = 0.000853

[H+] = x = 0.000853M

PH = -log[H+]

= -log0.000853

= 3.069 >>>answer

queen_honey_blossom answered 2 years ago

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