Chapter 17, Problem 96AP

Water containing ${\text{Ca}}^{\text{2+}}{\text{and Mg}}^{\text{2+}}$ ions is called hard water and is unsuitable for some household and industrial use because these ions react with soap to form insoluble salts, or curds. One way to remove the ${\text{Ca}}^{\text{2+}}$ ions from hard water is by adding washing soda ${\text{(Na}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{- 10H}}_{\text{3}}\text{O)}$. (a) The molar solubility of ${\text{CaCO}}_{\text{3}}\text{ is 9}\text{.3 }\times {\text{ 10}}^{\text{-5}}$ if. What is its molar solubility in a 0.050 M ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution? (b) Why are ${\text{Mg}}^{\text{2+}}$ ions not removed by this procedure? (c) The ${\text{Mg}}^{\text{2+}}$ ions are removed as $\text{Mg}{\left(\text{OH}\right)}_2$ by adding slaked lime $\left[\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\right]$ to the water to produce a saturated solution. Calculate the pH of a saturated Ca(OH), solution. (d) What is the concentration of ${\text{Mg}}^{\text{2+}}$ ions at this pH? (e) In general, which ion $\left({\text{Ca}}^{\text{2+}}{\text{or Mg}}^{\text{2+}}\right)$ would you remove first? Why?

Interpretation Introduction

Interpretation: The molar solubility of ${\text{CaCO}}_3$ in $0.050{\text{ M Na}}_2{\text{CO}}_3$ solution is to be calculated. The reason behind the non-removal of ${\text{Mg}}^{2+}$ by this procedure is to be explained. The pH of the saturated $\text{Ca}{\left(\text{OH}\right)}_2$ is to be calculated. The concentration of ${\text{Mg}}^{2+}$ ions in the calculated pH is to be determined. The ion that will be removedout of the given ions is to be determined.

Concept introduction:

When a compound that contains an ion that is common with a dissolved substance is added to a solution at equilibrium, the equilibrium shifts to the left. This is known as the common ion effect.

Molar solubility is defined as the moles of solute dissolved in one liter of solution, and its unit is $\text{mol}/\text{L}$.

At a given temperature, the product of molar concentrations of the ions of salt present in the solution is known as the solubility product of the salt. It is represented by $K_{sp}$.

Higher is the value of solubility product of a salt, more is its solubility.

The presence of common ions in the solution decreases the solubility of a given compound.

For a general reaction: $\text{AB}\left(s\right)\rightleftharpoons {\text{A}}^{+}\left(aq\right)+{\text{B}}^{-}\left(aq\right)$.

The solubility product can be calculated by the expression as: ${\text{K}}_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{][B}}^{-}\text{]}$

Here, ${\text{K}}_{\text{sp}}$ is the solubility product constant and $sp$ stands for solubility product.

The expression of $\text{pOH}$ is represented as:

$\text{pOH}=-\text{log}\left[\text{OH}\right]$

The relation between $\text{pH}$ and $\text{pOH}$ can be expressed as: $\text{pH}+\text{pOH}=14$

Answer to Problem 96AP

Solution:

(a) $1.7\times {10}^{-7}\text{ M}$

(b)It is fairly soluble.

(c)12.40

(d) $1.9\times {10}^{-8}\text{ M}$

(e) ${\text{Ca}}^{2+}$

Explanation of Solution

a)The molar solubility of ${\text{CaCO}}_3$ is $9.3\times {10}^{-5}\text{ M}$, it’s molar solubility in $0.050\text{ M}$ ${\text{Na}}_2{\text{CO}}_3$.

This problem is based on the common ion effect.

Consider the dissociation of ${\text{Na}}_2{\text{CO}}_3$ as:

${\text{Na}}_2{\text{CO}}_3\left(s\right)\to 2{\text{Na}}^{+}\left(aq\right)+{\text{CO}}_3{}^{2-}\left(aq\right)$

Let $s$ be the solubility of ${\text{CaCO}}_3$ in the ${\text{Na}}_2{\text{CO}}_3$ solution.

Consider the dissociation of ${\text{CaCO}}_3$ as:

$\begin{array}{l}{\text{ CaCO}}_3\left(s\right)\to {\text{Ca}}^{2+}\left(aq\right)+{\text{CO}}_3{}^{2-}\left(aq\right)\\ \text{Initial}\left(\text{M}\right)\text{ 0}\text{.00 0}\text{.050}\\ \text{Change}\left(\text{M}\right)\text{ +}s\text{ +}s\\ \text{Equilibrium}\left(\text{M}\right)s\text{ 0}\text{.050}+s\end{array}$

According to the above dissociation reaction, the expression for $K_{\text{sp}}$ will be as:

$K_{\text{sp}}=\left[{\text{Ca}}^{2+}\right]\left[{\text{CO}}_3{}^{2-}\right]$

The solubility product constant of ${\text{CaCO}}_3$ is $8.7\times {10}^{-9}$.

Substitute $8.7\times {10}^{-9}$ for $K_{\text{sp}}$, $s$ for $\left[{\text{Ca}}^{2+}\right]$, and $\left(0.050+s\right)$ for $\left[{\text{CO}}_3^{2-}\right]$ in the above expression as:

$8.7\times {10}^{-9}=s\left(0.050+s\right)$

Since the value of $K_{\text{sp}}$ is very small, the value of $\left(0.050+s\right)$ will be approximately equal to 0.050.

$\begin{array}{c}8.7\times {10}^{-9}=0.050s\\ s=\frac{8.7\times {10}^{-9}}{0.050}\\ =1.7\times {10}^{-7}\text{ M}\end{array}$

As the addition of washing soda to permanent hard water is done, most of the ${\text{Ca}}^{2+}$ will be removed due to the common ion effect.

Therefore, the molar solubility of ${\text{CaCO}}_3$ in $0.050{\text{ M Na}}_2{\text{CO}}_3$ solution is $1.7\times {10}^{-7}\text{ M}$.

b) ${\text{Mg}}^{2+}$ Ions not removed by this procedure

The solubility product constant of ${\text{MgCO}}_3$ is $4.0\times {10}^{-5}$. From its $K_{\text{sp}}$ value, it can be said that it is soluble in water. As a result, ${\text{Mg}}^{2+}$ cannot be removed by the common ion effect method.

c) The $\text{pH}$ of a saturated $\text{Ca}{\left(\text{OH}\right)}_2$ solution.

The solubility product constant of $\text{Ca}{\left(\text{OH}\right)}_2$ is $8.0\times {10}^{-6}$. Write its dissociation reaction as:

$\text{Ca}{\left(\text{OH}\right)}_2\to {\text{Ca}}^{2+}+2{\text{OH}}^{-}$

Let $s$ be the solubility of $\text{Ca}{\left(\text{OH}\right)}_2$.

So, according to the above reaction, the expression for $K_{\text{sp}}$ will be as:

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(2s\right)}^2\\ =4s^3\end{array}$

Substitute $8.0\times {10}^{-6}$ for $K_{\text{sp}}$ in the above expression as:

$\begin{array}{c}8.0\times {10}^{-6}=4s^3\\ s^3=\frac{8.0\times {10}^{-6}}{4}\\ =2.0\times {10}^{-6}\\ s={\left(2.0\times {10}^{-6}\right)}^{1/3}\\ =0.0126\text{ M}\end{array}$

Now, the concentration of ${\text{OH}}^{-}$ will be as:

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=2s\\ =2\left(0.0126\text{ M}\right)\\ =0.0252\text{ M}\end{array}$

Calculate the $\text{pOH}$ of the solution as:

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Substitute $0.0252\text{ M}$ for $\left[{\text{OH}}^{-}\right]$ as:

$\begin{array}{c}\text{pOH}=-\log \left[0.0252\text{ M}\right]\\ =1.60\end{array}$

The relation between $\text{pH}$ and $\text{pOH}$ is:

$\text{pH}+\text{pOH}=14$

Substitute 1.60 for $\text{pOH}$ as:

$\begin{array}{c}\text{pH}+1.60=14\\ \text{pH}=12.40\end{array}$

Therefore, the $\text{pH}$ of the $\text{Ca}{\left(\text{OH}\right)}_2$ solution is 12.40.

d)The concentration of ${\text{Mg}}^{2+}$ ions at this $\text{pH}$

From the previous part of the question, the concentration of ${\text{OH}}^{-}$ is $0.0252\text{ M}$, which is quite high. At this concentration, most of the ${\text{Mg}}^{2+}$ will be removed in the form of $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$. The concentration of the remaining ${\text{Mg}}^{2+}$ is only due to following reaction, which takes place at equilibrium:

$\text{Mg}{\left(\text{OH}\right)}_2\to {\text{Mg}}^{2+}+2{\text{OH}}^{-}$

According to the above reaction, the expression for $K_{\text{sp}}$ will be as:

$K_{\text{sp}}=\left[{\text{Mg}}^{2+}\right]{\left[{\text{OH}}^{-}\right]}^2$

The calculated concentration of ${\text{OH}}^{-}$ is $0.0252\text{ M}$.

Substitute $0.0252\text{ M}$ for $\left[{\text{OH}}^{-}\right]$ and $1.2\times {10}^{-11}$ for $K_{\text{sp}}$ in the above reaction as:

$\begin{array}{c}1.2\times {10}^{-11}=\left[{\text{Mg}}^{2+}\right]{\left[0.0252\text{ M}\right]}^2\\ \left[{\text{Mg}}^{2+}\right]=\frac{1.2\times {10}^{-11}}{{\left(0.0252\text{ M}\right)}^2}\\ =1.9\times {10}^{-8}\text{ M}\end{array}$

Therefore, the concentration of ${\text{Mg}}^{2+}$ is $1.9\times {10}^{-8}\text{ M}$.

e) The i on $\left({\text{Ca}}^{2+}{\text{ or Mg}}^{2+}\right)$ that would be removed first.

Out of ${\text{Ca}}^{2+}{\text{ \& Mg}}^{2+}$, ${\text{Ca}}^{2+}$ will be removed first because it is present in alarger quantity.