Question

A solution is prepared by dissolving 25.0 g of ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00-ml sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

Short answer

The concentration of ammonium ions and sulfate ions in the final solutionare $\left[N{H}_4^{+}\right]=0.630M$and $\left[S{O}_4^{2-}\right]=0.315M$.

Step-by-step solution

Definition

A solution's molarity is a concept that refers to the amount of solute contained in one litre of the solution. It may be stated mathematically as follows:

$Molarity=\frac{molesofsolute}{volumeofsolutioninliter}$

Determination of molarity of  NH42SO4

The first step is to compute the ammonium sulfate's initial molarity as follows:

$$\begin{gathered} molesof{\left(N{H}_4\right)}_{2}S{O}_4=\frac{Molesof{\left(N{H}_4\right)}_{2}S{O}_4}{Molarmassof{\left(N{H}_4\right)}_{2}S{O}_4} \\ =\frac{25.0g}{2(14.01)+8(1.008)+1(32.07)+4(16.00){\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}} \\ =0.189mol \end{gathered}$$

The molarity of ammonium sulphate solution must then be calculated

$$\begin{gathered} Molarityof{\left(N{H}_4\right)}_{2}S{O}_4=\frac{0.189mol}{100.0mL}\times \frac{1000mL}{1.0L} \\ =1.89M \end{gathered}$$

Determination of final molarity of  NH42SO4

The next step is to compute the ammonium sulphate final concentration after dilution. We may construct the following equation if $M_i$ is the initial molar concentration of the solution, $V_i$is the starting volume of the solution,role="math" localid="1650284396254" $M_f$is the final molar concentration, and $V_f$is the final volume of the solution.

$M_i{V}_i=M_f{V}_f$

We know that the Ammonium sulphatesolution has the following properties,

$$\begin{gathered} M_i=1.89M \\ {V}_i=10.00mL \\ {M}_f=? \\ {V}_f=\left(10.00+50.00\right)mL \\ {V}_f=60.00mL \end{gathered}$$

$1.89M\times 10.00mL=M_f\times 60.00mL$

$$\begin{gathered} M_f=\frac{1.89M\times 10.00mL}{60.00mL} \\ {M}_f=0.315M \end{gathered}$$

Thereforethe final morality of ${\left(N{H}_4\right)}_{2}S{O}_4$ is$0.315M$.

Determination of concentration of ammonium ions and sulfate ions

Let’s compute the concentration of ammonium ions and sulfate ions using molar ratio,

$$\begin{gathered} MolarityofN{H}_4^{+}=0.315M{\left(N{H}_4\right)}_{2}S{O}_4\times \frac{2molN{H}_4^{+}}{2mol{\left(N{H}_4\right)}_{2}S{O}_4} \\ =0.630MN{H}_4^{+} \\ MolarityofS{O}_4^{2-}=0.315M{\left(N{H}_4\right)}_{2}S{O}_4\times \frac{\text{1}molS{O}_4^{2-}}{1mol{\left(N{H}_4\right)}_{2}S{O}_4} \\ =0.315MS{O}_4^{2-} \end{gathered}$$

Therefore the concentration of ions are $\left[N{H}_4^{+}\right]=0.630M$and$\left[S{O}_4^{2-}\right]=0.315M$.