Answer to Question #67634 in General Chemistry for Natalie

Question

Ammonia gas, Kb = 1.8 x 10^-5, is bubbled into a solution of 250 mL of 0.050 M nitric acid until the pH of the solution rises to 9.26. How many milliliters of ammonia gas, measured at 25 C and 740 torr must me dissolved in the solution to observe this change in pH?

Answer 1

NH₃(aq)+HNO₃(aq)→NH₄NO₃(aq)
pOH= 14-9.26 = 4.74
pOH= pKb+lg([NH₄⁺]/[NH₃])
4.74= 4.744727495+lg([NH₄⁺]/[NH₃])
lg([NH₄⁺]/[NH₃])= 4.74-4.744727495 = -4.7274949x10⁻³
[NH₄⁺]/[NH₃]= 10^(-4.7274949x10⁻³) = 0.989173572
Number of moles of NH₄⁺/number of moles of NH₃
= 0.989173572
Number of moles of HNO₃ originally presents
= ([HNO₃])(original volume of the solution in L)
= (0.050)(250x10⁻³) = 0.0125
Let x be the number of moles of NH₃ added.
Number of moles of NH₃ = x-0.0125
Number of moles of NH₄NO₃ = 0.0125
0.0125/(x-0.0125)= 0.989173572
0.989173572x= 0.024864669
x= 0.025136811
We use the ideal gas equation, PV= nRT
760 torr= 1.01x10⁵Pa
740 torr= (1.01x10⁵/760)(740) = 9.834210526x10⁴Pa
25°C = (25+273)K = 298K
V= nRT/P
V = (0.025136811)(8.31)(298)/9.834210526x10⁴ = 6.329770535x10⁻⁴m³
1m³= 1x10⁶mL
6.329770535x10⁻⁴m³= (1x10⁶)(6.329770535x10⁻⁴) = 632.9770535mL = 633mL
Answer: 633mL