Question

Calculate the pH of 1.0 L upon addition of 0.190 mol of solid NaOH to the original buffer solution. Express the pH to two decimal places. Consider a buffer solution that is 0.50 M in NH3 and 0.20 M in NH4Cl. For ammonia, pKb=4.75.

Answer 1

Given,

The volume of buffer solution = 1.0 L

Moles of NaOH = 0.190 mol

Buffer solution with

[NH₃]= 0.50 M

[NH₄Cl] = 0.20 M

Calculating the number of moles of NH₃ and NH₄Cl,

Moles of NH₃ = 0.50 M x 1.0 L = 0.50 mol of NH₃

Similarly,

Moles of NH₄Cl Or NH₄⁺ = 0.20 M x 1.0 L = 0.20 mol of NH₄⁺

Now, the reaction between NH₄⁺ and NaOH is,

NH₄⁺(aq) + OH^-(aq) NH₃(aq) + H₂O(l)

Drawing an ICE chart,

	NH4+(aq)	OH-(aq)	NH3(aq)
I(moles)	0.20	0.190	0.50
C(moles)	-0.190	-0.190	+0.190
E(moles)	0.01	0	0.69

Now, the new concentrations of NH₄⁺ and NH₃ are,

[NH₄⁺] = 0.01 mol / 1 L = 0.01 M

[NH₃] = 0.69 mol / 1 L = 0.69 M

The equilbrium reaction for buffer is,

NH₃(aq) + H₂O(l) NH₄⁺(aq) + OH^-(aq)

Now, we know, the Henderson-Hasselbalch equation,

pOH = pK_b + log [ C.acid/ base]

pOH = 4.75 + log [ NH₄⁺/ NH₃]

pOH = 4.75 + log [0.01/ 0.69]

pOH = 4.75 - 1.84

pOH = 2.91

Now, we know,

pH + pOH = 14

Rearranging the formula,

pH = 14 - pOH

pH = 14 -2.91

pH = 11.09