Question

Sodium tripolyphosphate $\mathbf{\left(}{\mathbf{\text{Na}}}_{\mathbf{\text{5}}}{\mathbf{\text{P}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}_{\mathbf{\text{10}}}\mathbf{\right)}$is used in many synthetic detergents. Its major effect is to soften the water by complexing role="math" localid="1663829263651" ${\mathbf{\text{Mg}}}^{\mathbf{\text{2+}}}{\mathbf{\text{andCa}}}^{\mathbf{\text{2+}}}$ions. It also increases the efficiency of surfactants, or wetting agents that lower a liquid’s surface tension. The pK value for the formation of ${\mathbf{\text{MgP}}}_{\mathbf{\text{3}}}{{\mathbf{\text{O}}}_{\mathbf{\text{10}}}}^{\mathbf{\text{3-}}}$is role="math" localid="1663830233340" $\mathbf{\text{-8.60}}$. The reaction is ${\mathbf{\text{Mg}}}^{\mathbf{\text{2+}}}\mathbf{+}{\mathbf{\text{P}}}_{\mathbf{\text{3}}}{{\mathbf{\text{O}}}_{\mathbf{\text{10}}}}^{\mathbf{\text{5-}}}\mathbf{\rightleftarrows }{\mathbf{\text{MgP}}}_{\mathbf{\text{3}}}{{\mathbf{\text{O}}}_{\mathbf{\text{10}}}}^{\mathbf{\text{3-}}}$. Calculate the concentration of ${\mathbf{\text{Mg}}}^{\mathbf{\text{2+}}}$in a solution that was originally $\mathbf{50}$. $\mathbf{\text{ppm}}$($\mathbf{\text{50.mg}}\mathbf{/}\mathbf{\text{L}}$of solution) after $\mathbf{\text{40.g}}$of ${\mathbf{\text{Na}}}_{\mathbf{\text{5}}}{\mathbf{\text{P}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}_{\mathbf{\text{10}}}$is added to $\mathbf{\text{1.0L}}$ of the solution.

Short answer

Equilibrium concentration of magnesium ions is $1\times {10}^{-5}\text{M}$.

Step-by-step solution

Definition of synthetic detergents:

A synthetic detergent is a cleansing agent that works in the same way as soap but is created from chemical compounds rather than lipids and lye.

The given data:

Consider the given values as below.

The pK value for the formation of${\text{MgP}}_{\text{3}}{{\text{O}}_{\text{10}}}^{\text{3-}}$ is$-8.60$.

$$\begin{gathered} \text{λ}\left({\text{Mg}}^{\text{2+}}\right)=50\raisebox{1ex}{$\text{mg}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right. \\ \text{λ}\left({\text{Na}}_{\text{5}}{\text{P}}_{\text{3}}{\text{O}}_{\text{10}}\right)=40\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right. \end{gathered}$$

Calculating molar concentration out of mass concentration:

Simplifying further,

$\lambda =\frac{m}{V}$

Here, $m=n\times M_r$

Therefore,

$\begin{array}{rcl}\lambda & =& \frac{n\times M_r}{V}\\ & =& c\times M_r\end{array}$

$\begin{array}{rcl}{c}_1& =& \frac{\lambda \left(M{g}^{2+}\right)}{A_{\Gamma }\left(M{g}^{2+}\right)}\\ & =& \frac{50\times {10}^{-3}{\displaystyle }{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.}}{24.31{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& 2.06\times {10}^{-3}\text{M}\end{array}$

Obtaining the concentration:

Let us find the required value of concentration,

$\begin{array}{rcl}{\text{M}}_{\text{r}}\left({\text{Na}}_{\text{5}}{\text{P}}_{\text{3}}{\text{O}}_{\text{10}}\right)& =& 5\times {\text{A}}_{\text{r}}\left(\text{Na}\right)+3\times A_r\left(\text{P}\right)+10\times A_r\left(\text{O}\right)\\ & =& 5\times \left(22.99\frac{\text{g}}{\text{mol}}\right)+3\times \left(30.97\frac{\text{g}}{\text{mol}}\right)+10\times \left(16\frac{\text{g}}{\text{mol}}\right)\\ & =& 357.86\frac{\text{g}}{\text{mol}}\end{array}$

$\begin{array}{rcl}{c}_2& =& \frac{\gamma \left({\text{Na}}_{\text{5}}{\text{P}}_{\text{3}}{\text{O}}_{\text{10}}\right)}{M_r\left({\text{Na}}_{\text{5}}{\text{P}}_{\text{3}}{\text{O}}_{\text{10}}\right)}\\ & =& \frac{40{\displaystyle \frac{\text{g}}{\text{L}}}}{367.86{\displaystyle \frac{\text{g}}{\text{mol}}}}\\ & =& 0.11\text{M}\end{array}$

Calculating k value:

Now, you will calculate constant $\left(\text{K}\right)$and place an equation for constant $\left(\text{K}\right)$using reactants and products Equilibrium concentrations.

$\text{pK}=-\text{log(K)}$

$\begin{array}{rcl}\text{K}& =& {\text{10}}^{\text{-pK}}\\ & =& {10}^{-(-8.60)}\\ & =& 3.98\times {10}^8\\ & =& \frac{\left[{\text{MgP}}_{\text{3}}{\text{O}}_{\text{10}}^{\text{3}-}\right]}{\left[{\text{Mg}}^{\text{2}+}\right]\left[{\text{P}}_{\text{3}}{\text{O}}_{\text{10}}^{\text{3}-}\right]}\end{array}$

Calculating the equilibrium concentration of product:

The equilibrium concentration of product in the numerator will be marked with $x$, each reactant will be written if form initial concentration minus $x$because for each spent reactant, one product will be formed.

$\begin{array}{rcl}K& =& \frac{x}{(c_1-x)(c_2-x)}\\ & =& \frac{x}{c_1{c}_2-c_{1}x-c_{2}x+x^2}\end{array}$

role="math" localid="1663832132968" $$\begin{gathered} K{c}_1{c}_2-xK{c}_1-xK{c}_2+x^{2}K-x=0 \\ {x}_{1,2}=\frac{K{c}_1+K{c}_2+1-/+\sqrt{{(K{c}_1+K{c}_2+1)}^2-4KK{c}_1{c}_2}}{2K} \\ x=2.05\times {10}^{-3}M \end{gathered}$$

Calculating the equilibrium of magnesium:

The final step is to calculate equilibrium magnesium concentration in the solution using initial concentration and spent concentration $x$.

$\begin{array}{rcl}[{\text{Mg}}^{\text{2+}}& \mid & =c_1-xn\\ & =& 2.06\times {10}^{-3}\text{M}-2.05\times {10}^{-3}\text{M}\\ & =& 1\times {10}^{-5}\text{M}\\ & & \end{array}$

The solutions in the book indicate that the final concentration is incorrect, but a calculator is required in this case because a conventional calculator cannot be used to calculate it.

Hence, Equilibrium concentration of magnesium ions is $1\times {10}^{-5}\text{M}$.