Question

A solution is prepared to be 0.1475 M in $Sr{{\left( OH \right)}_{2}}$ . How do you find the $\left[ {{H}_{3}}{{O}^{+}} \right]$ , $\left[ O{{H}^{-}} \right]$ , pH, and pOH ?

Answer 1

Hint: In order to solve this question, we will first write the ${{K}_{b}}$ base dissociation constant then we will calculate the value of pH by using the formula of pH that is $pH=-\log \left[ {{H}_{3}}{{O}^{+}} \right]$ . Then will calculate the value of pOH by using the formula:
$pOH = -\log \left[ O{{H}^{-}} \right]$

Complete step by step answer:
- As we are being provided with the information that a solution is prepared to be 0.1475 M in $Sr{{\left( OH \right)}_{2}}$.
- As we know that Strontium Hydroxide ionizes in water to produce the hydroxide ion and strontium ion. It is found that strontium is slightly soluble in water.
- Now, we can write the reaction of dissociation of strontium as:
$Sr{{\left( OH \right)}_{2}}\rightleftarrows S{{r}^{2+}}+2O{{H}^{-}}$
- Here, we can see that strontium hydroxide is being dissociated into strontium ions and hydroxide ions.
$\begin{align}
  & Sr{{\left( OH \right)}_{2}}\rightleftarrows S{{r}^{2+}}+2O{{H}^{-}} \\
 & 0.1475\text{ - -} \\
 & -x\text{ +x +x} \\
 & 0.1475-x\text{ x 2x} \\
\end{align}$ - Now, we will write the value of${{K}_{b}}$ base dissociation constant as:
$\begin{align}
  & {{K}_{b}}=\dfrac{{{\left[ O{{H}^{-}} \right]}^{2}}\left[ S{{r}^{2+}} \right]}{\left[ Sr{{\left( OH \right)}_{2}} \right]} \\
 & {{K}_{b}}=\dfrac{{{\left[ 2x \right]}^{2}}\left[ x \right]}{\left[ 0.1475-x \right]} \\
 & =6.5\times {{10}^{-3}} \\
\end{align}$
Now, we will solve the value of x as:
x= 0.05346 M.
Now, we can write$\left[ O{{H}^{-}} \right]$ as:
$\left[ O{{H}^{-}} \right]=2x=2\times 0.05346=0.1069M$
- As we know that in any aqueous solution, both ions that are $\left[ O{{H}^{-}} \right]$ and $\left[ {{H}_{3}}{{O}^{+}} \right]$ are present and they must satisfy the following conditions:
$\begin{align}
 & \left[ O{{H}^{-}} \right]\left[ {{H}_{3}}{{O}^{+}} \right]={{K}_{w}} \\
 & \left[ O{{H}^{-}} \right]\left[ {{H}_{3}}{{O}^{+}} \right]=1.0\times {{10}^{-14}} \\
\end{align}$ - As we have determined the value of $\left[ O{{H}^{-}} \right]$ as 0.1.69 M. Now, further we can solve as:
$\begin{align}
 & \left[ {{H}_{3}}{{O}^{+}} \right]=\dfrac{1.0\times {{10}^{-14}}}{\left[ O{{H}^{-}} \right]} \\
 & \left[ {{H}_{3}}{{O}^{+}} \right]=\dfrac{1.0\times {{10}^{-14}}}{\left[ 0.1069 \right]} \\
 & \left[ {{H}_{3}}{{O}^{+}} \right]={{9.355}^{-14}}M \\
\end{align}$ So, we get the value of $\left[ {{H}_{3}}{{O}^{+}} \right]$ as ${{9.355}^{-14}}M$ .
- Now, we will calculate the value of pH by using the formula of pH as:
$\begin{align}
  & pH=-\log \left[ {{H}_{3}}{{O}^{+}} \right] \\
 & pH=-\log \left[ {{3.955}^{-14}} \right] \\
 & pH=13.0290 \\
\end{align}$
- We can calculate the value of pOH by using the formula:
$pOH=-\log \left[ O{{H}^{-}} \right]$
On calculating we get:
$\begin{align}
  & pOH=-\log \left[ 0.1069 \right] \\
 & pOH=0.9710 \\
\end{align}$ - Hence, we get the value of pH, and pOH as 13.0290 and 0.9710.

Note: - We should not get confused in the symbols ${{K}_{b}}$ and ${{K}_{B}}$ . As ${{K}_{b}}$ is a base dissociation constant that basically measures how completely a base dissociates into its component ions in water.
- Whereas, ${{K}_{B}}$ is the Boltzmann constant which relates the average kinetic energy of the particles present in a gas with the temperature of the gas.