Question

1.475-g sample containing ${\mathbf{NH}}_{\mathbf{4}}\mathbf{Cl}\mathbf{(}\mathbf{FM}\mathbf{53}\mathbf{.}\mathbf{491}\mathbf{)}\mathbf{,}{\mathbf{K}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$(FM 138.21), and inert ingredients was dissolved to give 0.100 L of solution. A 25.0-mL aliquot was acidified and treated with excess sodium tetraphenylborate,${\mathbf{Na}}^{\mathbf{+}}\mathbf{B}\mathbf{(}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{)}}_{\mathbf{4}}^{\mathbf{-}}$, to precipitate${\mathbf{K}}^{\mathbf{+}}$and

${\mathbf{NH}}_{\mathbf{4}}^{\mathbf{+}}$ions completely:

$$\begin{gathered} \mathbf{(}{\mathit{C}}_{\mathbf{6}}{\mathit{H}}_{\mathbf{5}}{\mathbf{)}}_{\mathbf{4}}{\mathit{B}}^{\mathbf{-}}\mathbf{+}{\mathit{K}}^{\mathbf{+}}\mathbf{\to }\mathbf{(}{\mathit{C}}_{\mathbf{6}}{\mathit{H}}_{\mathbf{5}}{\mathbf{)}}_{\mathbf{4}}\mathit{B}\mathit{K}\mathbf{\left(}\mathit{s}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{F}\mathit{M}\mathbf{358}\mathbf{.}\mathbf{33} \\ \mathbf{(}{\mathit{C}}_{\mathbf{6}}{\mathit{H}}_{\mathbf{5}}{\mathbf{)}}_{\mathbf{4}}{\mathit{B}}^{\mathbf{-}}\mathbf{+}\mathit{N}{\mathit{H}}_{\mathbf{4}}^{\mathbf{+}}\mathbf{\to }\mathbf{(}{\mathit{C}}_{\mathbf{6}}{\mathit{H}}_{\mathbf{5}}{\mathbf{)}}_{\mathbf{4}}\mathit{B}\mathit{N}{\mathit{H}}_{\mathbf{4}}\mathbf{\left(}\mathit{s}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{F}\mathit{M}\mathbf{337}\mathbf{.}\mathbf{27} \end{gathered}$$

The resulting precipitate amounted to 0.617 g. A fresh 50.0-mL aliquot of the original solution was made alkaline and heated to drive off all the${\mathbf{NH}}_{\mathbf{3}}\mathbf{.}$

$\mathit{N}{\mathit{H}}_{\mathbf{4}}^{\mathbf{+}}\mathbf{+}\mathit{O}{\mathit{H}}^{\mathbf{-}}\mathbf{\to }\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}$

It was then acidified and treated with sodium tetraphenylborate to give 0.554 g of precipitate. Find the weight percent of${\mathbf{NH}}_{\mathbf{4}}\mathbf{Cl}$and${\mathbf{K}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$in the original solid.

Short answer

The weight percentage of$\mathrm{m}\left({\mathrm{NH}}_4\mathrm{Cl}\right)$ is 14.6%

The weight percentage of $\mathrm{m}\left({\mathrm{N}}_2{\mathrm{CO}}_3\right)$is 14.5%

Step-by-step solution

Calculate x and y:

Let consider,

X as $\mathrm{m}\left({\mathrm{NH}}_4\mathrm{Cl}\right)$and y as $m\left(K_{2}C{O}_3\right)$

25 ml of the sample gave 0.617 g of precipitate which contained both of the products

$$\begin{gathered} \frac{1}{4}\left(\frac{x}{M\left(N{H}_{4}Cl\right)}\right)\times M\left(\right(C_6{H}_5{)}_{4}BN{H}_4)+\left(\frac{\left(2y\right)}{M\left(K_{2}C{O}_3\right)}\times M\left(\right(C_6{H}_5{)}_{4}BK)\right)=0.617g \\ \frac{1}{4}\left(\frac{x}{53.492}\times 337.27+\frac{2y}{138.21}\times 358.33\right)=0.617g \end{gathered}$$

Calculate for y we get,

$$\begin{gathered} \frac{1}{2}\left(\frac{2y}{M\left(K_{2}C{O}_3\right)}\times M\left(\right(C_6{H}_5{)}_{4}BK\left)\right)\right)=0.554g \\ \frac{1}{2}\left(\frac{2y}{138.21}\times 358.33\right)=0.554g \\ y=0.2137g \end{gathered}$$

Calculate for x we get,

$$\begin{gathered} \frac{1}{4}\left(\frac{x}{53.492}\times 337.27+\frac{2\times 0.2137g}{138.21}\times 358.33\right)=0.617g \\ x=0.2157g \end{gathered}$$

Calculate the weight percentages:

$$\begin{gathered} wt\%\left(x\right)=\frac{x}{m(sample)} \\ wt\%\left(x\right)=\frac{0.2157g}{1.475g} \\ wt\%\left(x\right)=14.6\% \end{gathered}$$

The weight percentage of$\mathrm{m}\left({\mathrm{NH}}_4\mathrm{Cl}\right)$is 14.6%

role="math" localid="1663664775980" $$\begin{gathered} wt\%\left(y\right)=\frac{y}{m(sample)} \\ wt\%\left(y\right)=\frac{0.2137g}{1.475g} \\ wt\%\left(y\right)=14.5\% \end{gathered}$$

The weight percentage of $m\left(K_{2}C{O}_3\right)$is 14.5%