Question

23. A buffer consists of 0.65 M of the weak monoprotic base B and 0.50 M of it salt BHCl. What is the pH of the buffer after 0.006 moles of KOH are added to 0.20 L of this buffer solution? Kb for the monoprotic base B is 1.3× 10‒3.

Answer 1

Given buffer contain weak base B and its conjugate acid BH ⁺ . K _b of weak base is 1.3 $\times$ 10 ^-3.

We have relation, pK _b = - log K _b

$\therefore$  pK _b = - log ( 1.3 $\times$ 10 ^-3. ) = 2.89

Now, calculate concentration of B & BH ⁺ in moles.

We have, Molarity = No. of moles of solute / volume of solution in L

$\therefore$ No. of moles of solute = Molarity $\times$ volume of solution in L

$\therefore$ No. of moles of B = 0.65 mol / L $\times$ 0.20 L = 0.13 mol

$\therefore$ No. of moles of BH ⁺ = 0.5 mol / L $\times$ 0.20 L = 0.10 mol

Consider reaction of KOH with buffer solution. KOH reacts with BHCl and produces water and base B.

BHCl + KOH $\rightarrow$ B + H₂O + KCl

Let' use ICE table.

Concentration ( moles )	BHCl	KOH	B
I	0.10	0.006	0.13
C	- 0.006	-0.006	+0.006
E	0.094	0.000	0.136

We can calculate pOH of buffer solution by using Henderson's equation as shown below.

pOH = pK b + log [ salt] / [ Base ]

$\therefore$ pOH = 2.89 + log [ BHCl ] / [ B]

$\therefore$ pOH = 2.89 + log ( 0.094 mol / 0.20 L )  / ( 0.136 mol / 0.20 L )

p OH = 2.89 - 0.1604

p OH = 2.73

We have relation, p H + p OH = 14

$\therefore$ p H = 14 - p OH = 14 - 2.73 = 11.27

ANSWER : pH of buffer solution after addition of 0.006 mol KOH = 11.27