Question

What is the pH of 1.00L of an ammonia/ammonium, NHz/NH4+, base buffer solution containing 0.35M NH; and 0.35M NH,* after 0.030 moles of the strong acid, HNO3, are added (NHz Ky - 1.8 x 10-$)? O pH = 5.2 - 7.2 O pH = 9.2 pH - 12

Answer 1

Concentration of NH3 = 0.35 M = 0.35 mol/L

Volume of NH3 = 1.00 L

Moles of NH3 = (0.35 mol/L)(1.00 L) = 0.35 mol

Concentration of NH4+ = 0.35 M = 0.35 mol/L

Volume of NH4+ = 1.00 L

Moles of NH4+ = (0.35 mol/L)(1.00 L) = 0.35 mol

Moles of HNO3 = 0.030 mol

Reaction	NH3(aq)	+	HNO3(aq)	$\rightarrow$	NH4+(aq)	+	NO3-(aq)
Initial Moles	0.35 mol		0.030 mol		0.35 mol		0.00 mol
Change in Moles	-0.030 mol		-0.030 mol		+0.030 mol		+0.030 mol
Final Moles	0.32 mol		0.00 mol		0.38 mol		0.030 mol

Moles of NH3 = 0.32 mol

Volume of NH3 = 1.00 L

Concentration of NH3 = (0.32 mol)/(1.00 L) = 0.32 mol/L = 0.32 M

Moles of NH4+ = 0.38 mol

Volume of NH4+ = 1.00 L

Concentration of NH4+ = (0.38 mol)/(1.00 L) = 0.38 mol/L = 0.38M

Reaction	NH3(aq)	+	H2O(l)	$\rightleftharpoons$	NH4+(aq)	+	OH-(aq)
Initial Concentration	0.32 M		PURE		0.38 M		0 M
Change in Concentration	-y M				+y M		+y M
Equilibrium Concentration	(0.32 - y) M		LIQUID		(0.38 + y) M		y M

Kb = [NH4+][OH-]/[NH3] = 1.8 x 10-5

(y)(0.38 + y)/(0.32 - y)] = 1.8 x 10-5

y2 + (0.38 + 1.8 x 10-5)y - (0.32 x 1.8 x 10-5) = 0

Solving this, we get,

y = 1.515 x 10-5 M

[OH-] = 1.515 x 10-5 M

Now, pOH = -log10([OH-]) = -log10(1.515 x 10-5) = 4.82

Also, pH = 14 - pOH = 14 - 4.82 = 9.18 $\approx$ 9.2

Therefore, pH of the solution is 9.2 and hence, the Correct Answer is Option (3).