Question

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 is dissolved in water to produce a 0.0141 M solution. The pH of the resulting solution is 2.50. Calculate the Ka for the acid.

Answer 1

1)

1)

Lets write the acid as HA

Lets write the dissociation equation of HA

HA -----> H+ + A-

0.3 0 0

0.3-x x x

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

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((4.5*10^-6)*0.3) = 1.162*10^-3

since c is much greater than x, our assumption is correct

so, x = 1.162*10^-3 M

so,

[H+] = x = 1.162*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (1.162*10^-3)

= 2.93

Answer: 2.93

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