The OH$$^-$$ concentration in a mixture of 5.0 mL of 0.0504 M NH$$_4$$Cl and 2 mL of 0.0210 M NH$$_3$$ solution is $$x \times 10^{-6}$$ M. The value of $$x$$ is _________ (Nearest integer) [Given $$K_w = 1 \times 10^{-14}$$ and $$K_b = 1.8 \times 10^{-5}$$]

We begin by noting that the mixture contains a weak base, ammonia NH$$_3$$, and its conjugate acid, the ammonium ion NH$$_4^+$$ (supplied by NH$$_4$$Cl). Such a pair constitutes a buffer system that obeys the equilibrium

$$\mathrm{NH_3\;+\;H_2O \rightleftharpoons NH_4^+ \;+\; OH^-}$$

For this equilibrium, the base-dissociation constant is given as

$$$K_b \;=\;\frac{[\mathrm{NH_4^+}]\,[\mathrm{OH^-}]}{[\mathrm{NH_3}]} \;=\;1.8\times10^{-5}$$$

Our first task is to calculate the molar concentrations of NH$$_3$$ and NH$$_4^+$$ after mixing the two solutions.

Number of moles of NH$$_3$$ present:

$$$n_{\mathrm{NH_3}} \;=\; V \times M \;=\; (2.0\;\text{mL})\;\Big(\frac{1\;\text{L}}{1000\;\text{mL}}\Big)\;\times 0.0210\;\text{M}$$$

$$$=\;0.0020\;\text{L}\times0.0210\;\text{mol L}^{-1} \;=\;4.20\times10^{-5}\;\text{mol}$$$

Number of moles of NH$$_4^+$$ (from NH$$_4$$Cl):

$$$n_{\mathrm{NH_4^+}} \;=\; V \times M \;=\; (5.0\;\text{mL})\;\Big(\frac{1\;\text{L}}{1000\;\text{mL}}\Big)\;\times 0.0504\;\text{M}$$$

$$$=\;0.0050\;\text{L}\times0.0504\;\text{mol L}^{-1} \;=\;2.52\times10^{-4}\;\text{mol}$$$

The total volume after mixing is simply the sum of the two volumes (volume additivity is assumed):

$$$V_{\text{total}} \;=\; 5.0\;\text{mL} \;+\; 2.0\;\text{mL} \;=\; 7.0\;\text{mL} \;=\; 0.0070\;\text{L}$$$

Hence the initial molarities (before any further equilibrium shift) are

$$$[\mathrm{NH_3}]_0 \;=\;\frac{4.20\times10^{-5}\;\text{mol}}{0.0070\;\text{L}} \;=\;6.00\times10^{-3}\;\text{M}$$$

$$$[\mathrm{NH_4^+}]_0 \;=\;\frac{2.52\times10^{-4}\;\text{mol}}{0.0070\;\text{L}} \;=\;3.60\times10^{-2}\;\text{M}$$$

Because NH$$_3$$ is weak and NH$$_4^+$$ is its conjugate acid, only a very small additional amount of NH$$_3$$ will convert to NH$$_4^+$$ and OH$$^-$$. We let the equilibrium concentration of OH$$^-$$ be $$y$$:

$$$\begin{aligned} \mathrm{NH_3} &\;\;+\;\; \mathrm{H_2O} &\rightleftharpoons&\;\; \mathrm{NH_4^+} &+;;& \mathrm{OH^-} \\ \text{Initial (M)} &\;6.00\times10^{-3} &&\;3.60\times10^{-2} && 0 \\ \text{Change (M)} &\;-y && +y && +y\\ \text{Equilibrium (M)} &6.00\times10^{-3}-y && 3.60\times10^{-2}+y && y \end{aligned}$$$

Because the buffer is fairly concentrated and $$y$$ will be very small compared with the starting concentrations, we make the common buffer approximation $$6.00\times10^{-3}-y \approx 6.00\times10^{-3}$$ and $$3.60\times10^{-2}+y \approx 3.60\times10^{-2}$$. Substituting these equilibrium values into the definition of $$K_b$$ gives

$$$K_b \;=\;\frac{(3.60\times10^{-2})\,y}{6.00\times10^{-3}}$$$

Solving for $$y$$ (which equals $$[\mathrm{OH^-}]$$):

$$$y \;=\;\frac{K_b\; \bigl(6.00\times10^{-3}\bigr)}{3.60\times10^{-2}}$$$

$$$=\;\frac{1.8\times10^{-5}\;\times\;6.00\times10^{-3}}{3.60\times10^{-2}}$$$

Now we simplify step by step:

$$$1.8\times10^{-5}\;\times\;6.00\times10^{-3} = 10.8\times10^{-8} = 1.08\times10^{-7}$$$

$$$\frac{1.08\times10^{-7}}{3.60\times10^{-2}} = \frac{1.08}{3.60}\times10^{-5} = 0.30\times10^{-5} = 3.0\times10^{-6}$$$

Thus

$$[\mathrm{OH^-}] \;=\;3.0\times10^{-6}\;\text{M}$$

The question states this concentration in the form $$x\times10^{-6}\;\text{M}$$, so clearly

$$x = 3$$

As 3 is already an integer, taking the nearest integer does not alter the value.

Hence, the correct answer is Option C.