Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0167 M solution. The pH of the resulting solution is 2.36. Calculate the Ka for the acid.

A certain weak base has a Kb of 7.50 × 10-7. What concentration of this base will produce a pH of 10.25?

Answer #1

First, use your pH to determine [H+] using 10^-pH=
10^-2.36=4.365x10^-3

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

However you can not disregard x in this case because it is close to
your final concentration.

So, you have (4.365x10^-3)^2/0.0167-(4.365x10^-3)

Ka=1.545x10^-3

Kb= 7.50 × 10-7

The equilibrium reaction is below

B + H2O <=> CA + HO-

Equilibrium equation:

Kb= [HO-][CA]/[B]

[B]= concentration of base [CA]= concentration of conjugate
acid

Also [HO-]= [CA] = x

14 - pH = pOH............................14-10.25 = 3.75

-log[HO-] = pOH................................... -log[HO-]=
3.75

[HO-] = 10^ (-pOH) .............................. [HO-] = 10^-3.75
= 1.778 x 10^-4

Since [HO-]= [CA] = x and we now know that x is 1.778 x 10^-4

So using the equilibrium equation and replacing x for [HO-] and
[CA] we get:

Kb = x^2/[B]

Solve for [B]

[B] = x^2/Kb

[B] = (1.778 x 10^-4)^2 / (7.5 × 10-7)

[B]= 0.0422

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