(Solved): Ammonia (NH3) in water is in equilibrium with ammonium (NH4+), according to the following equ ...

Ammonia $\left({\mathrm{NH}}_3\right)$ in water is in equilibrium with ammonium $\left({\mathrm{NH}}_4^{+}\right)$, according to the following equation: ${\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}?{\mathrm{NH}}_4{}^{+}+{\mathrm{OH}}^{?}{\mathrm{K}}_{\mathrm{N}}=1.82\times 10^{?5}$ a. Write the equilibrium equation for this reaction. ${\mathrm{K}}_{\mathrm{N}}=1.82\times 10^{?5}$ ${\mathrm{KH}}_3+{\mathrm{H}}_2\mathrm{O}?{\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{?}\times {\mathrm{KN}}_N=\frac{\text{ prod }}{\text{ reant }}$ b. What is the value of ${\mathrm{pK}}_{\mathrm{N}}$ ? $\begin{array}{l}{K}_n=1.82\times 10^{?5}\\ ??\log \left(1.82\times 10^{?5}\right)=4.74\end{array}$ c. Calculate the ratio $\left[{\mathrm{NH}}_3\right]/\left[{\mathrm{NH}}_4^{+}\right]$at $\mathrm{pH}10$. $\begin{array}{l}\text{ pH }+\text{ poH }=14\\ \text{ poH }=12?ph\\ \text{ polt }=14?10=4\\ ?\log \left[{\mathrm{gH}}_7\right]=10^{?4}\end{array}$ . If you wanted to shift the reaction to the right to decrease the ammonia concentration, would you raise the $\mathrm{pH}$, decrease the $\mathrm{pH}$, or remove water? Circle one and briefly explain.