Question

A solution of 0.025 M NH2OH (Ka-1.11 x 10") (20 pt) a) Calculate Kb b) Calculate the [OH). [NH,OH) and (NH,OH'). c) Calculate the pH and POH of the solution. d) What is the pH of the solution if the solution was mixed with 0.78 M NH3OH"?

Answer 1

a) Kb can be calculated from the relation:

$Ka\cdot Kb=10^{-14}$

So:

$Kb=\frac{10^{-14}}{Ka}=\frac{10^{-14}}{1.11x10^{-6}}=9.1x10^{-9}$

b) We need to build an ICE chart for the equilibrium:

$NH_{2}OH+H_{2}O\leftrightharpoons NH_{3}OH^{+}+OH^{-}$

Where the water concentration can be considered as constant:

	[NH2OH]	[NH3OH+]	OH-
initial	0.025 M	0	0
change	-x	+x	+x
equilibrium	0.025 M - x	x	x

$Kb=9.1x10^{-9}=\frac{[NH_{3}OH^{+}][OH^{-}]}{[NH_{2}OH]}=\frac{x^{2}}{0.025-x}$

This is a quadratic equation, which can be solved to yield:

$x_{1}=1.51x10^{-5}\: M;x_{2}=-1.51x10^{-5}\: M$

The positive result is the one we are looking for, which is equivalent to the concentrations of OH^- and of NH₃OH^-. The concentration of NH₂OH is:

$0.025\: M-1.51x10^{-5}\: M=0.024985\: M$

c) The pOH is given by:

$pOH=-log[OH^{-}]=-log[1.51x10^{-5}]=4.82$

And the pH:

$pH=14-pOH=14-4.82=9.18$

d) If there were also 0.78 M NH₃OH⁺, the pH could be calculated using the Henderson-Hasselbach equation:

$pH=pKa+log\left ( \frac{[NH_{2}OH]}{[NH_{3}OH^{+}]} \right )=-log(1.11x10^{-6})+log\left ( \frac{0.025M}{0.78M} \right )=4.46$