If the Ka of a monoprotic weak acid is 3.3 × 10-6, what is the pH of a 0.30 M solution of this acid?
let us consider a monoprotic weak acid HA , basic acid dissociates as follows:
HA --- --> H+ + A- , using ICE rule we can find the pH
Initially 0.30 0 0
change -x +x +x
at equilibrium (0.30-x) x x
now, ka=[H+][A-] /[HA]
=> 3.3 × 10-6 = X2 /(0.30-X) [As the value of ka is very small , so x in the denominator is neglected]
=>x2 = 3.3 × 10-6 * 0.30
Using quadratic formula , with a=1, b=0, c=-9.899999999999998e-7 , we get
so , [H+] = 0.0009M
We know , pH =-log [H+]
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