Question

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?

Answer #1

Ans---

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} = X^{2} /(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

=>x2=0.3(3.3(10−6)

Using quadratic formula , with a=1, b=0, c=-9.899999999999998e-7 , we get

x=0.00099

so , [H+] = 0.0009M

We know , pH =-log [H+]

pH=-log(0.0009)

pH=3.04

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1.) If the Ka of a monoprotic
weak acid is 4.5 x 10^-6, what is the pH of a 0.30M solution of
this acid?
2.) The Ka of a monoprotic weak
acid is 7.93 x 10^-3. What is the percent ionization of a 0.170 M
solution of this acid?
3.) Enolugh of a monoprotic acid
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a. Find the pH of a 0.100 M solution of a weak
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b.Find the percent dissociation of this solution.
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d. Find the percent
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f.Find the percent dissociation of this solution.

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