Question

A 1.0-L buffer solution initially contains 0.25 mol of NH3 (Kb=1.76×10−5) and 0.25 mol of NH4Cl.

In order to adjust the buffer pH to 8.85, should you add NaOH or HCl to the buffer mixture?

What mass of the correct reagent should you add?

Answer 1

First we need to calculate pH of original buffer solution.

pH of basic buffer solution is calculated by using Henderson's equation.

pOH = pKb + log [NH₄Cl] / [NH₃]

pOH = - log (1.76 $\times$ 10 ^-05 ) + log ( 0.25 mol / 1 L ) / ( 0.25 mol / 1 L )

pOH = 4.75 + 0

pOH = 4.75

We have, pH + pOH = 14

Therefore, pH = 14 - pOH = 14 - 4.75 = 9.25

pH of original buffer solution.

We have to adjust pH of buffer solution to 8.85. This value is less than original pH value (9.25) i e we have to increase acidity of buffer solution .To increase acidity of buffer solution ,we need to add HCl to above buffer.

Now consider reaction of HCl with buffer solution.

NH₃ + HCl $\rightarrow$ NH₄Cl

Let's use ICE table

Concentration ( moles)	NH3	HCl	NH4Cl
Initial	0.25	X	0.25
Change	- X	-X	+X
Equilibrium	0.25- X	0.00	0.25 + X

$\therefore$ pOH = 4.75 + log 0.25 + X / 0.25- X

( 14 - 8.85 ) = 4.75 + log 0.25 + X / 0.25- X ( $\because$ pH + pOH = 14 hence, pOH = 14 -pH)

5.15 = 4.75 + log 0.25 + X / 0.25- X

log 0.25 + X / 0.25- X = 5.15 - 4.75 =0.40

(0.25 + X / 0.25- X) = 2.51

0.25 + X =( 0.25 -X ) 2.51

0.25 + X = 0.628 - 2.51 X

X + 2.51 X = 0.628 - 0.25

3.51 X = 0.378

X = 0.378 / 3.51

X = 0.108 mol = moles of HCl

We have, No. of moles of HCl = Mass / Molar Mass

Mass of HCl = No. of moles of HCl $\times$ Molar mass of HCl

Mass of HCl = 0.108 mol $\times$ 36.46 g /mol

=3.94 g

ANSWER : To adjust pH of buffer solution to 8.85 we need to add 3.94 g HCl.