Answer to Question #143791 in General Chemistry for Add

Question

(15’) How many ml of 2.00 M HNO3 would be required to neutralize 12.5 ml of 0.0800 M NH3? What is the pH of the resulting solution? (KaNH4=5.6x10-10)

Answer 1

NH₃ + HNO₃ = NH₄NO₃

CaVa/CbVb = na/nb

2×Va/ 0.08× 12.5 = 1/1

Va= 0.08×12.5/2

Va= 0.5mL

Mole of HNO3 and NH3= mole of NH4NO3

mole of HNO3 = 2×0.5×10^-3 = 1×10^-3 mol

Concentration of NH4NO3 = 1×10^-3mol/[(0.5+12.5) × 10^-3 L]

= 0.0769M

PH of solution= 1/2 Pka - 1/2 log c

= 1/2(-log 5.6×10^-10 )- 1/2 log 0.0769

=5.18