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Question: A solution is prepared to be 0.1475 M in \(Sr{{\left( OH \right)}_{2}}\) . How do you find the \(\le...

A solution is prepared to be 0.1475 M in Sr(OH)2Sr{{\left( OH \right)}_{2}} . How do you find the [H3O+]\left[ {{H}_{3}}{{O}^{+}} \right] , [OH−]\left[ O{{H}^{-}} \right] , pH, and pOH ?

Explanation

Solution

In order to solve this question, we will first write the Kb{{K}_{b}} base dissociation constant then we will calculate the value of pH by using the formula of pH that is pH=−log⁡[H3O+]pH=-\log \left[ {{H}_{3}}{{O}^{+}} \right] . Then will calculate the value of pOH by using the formula:
pOH=−log⁡[OH−]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(OH)2Sr{{\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(OH)2⇄Sr2++2OH−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.
Sr(OH)2⇄Sr2++2OH− 0.1475 - - −x +x +x 0.1475−x x 2x \begin{aligned} & 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{aligned} - Now, we will write the value ofKb{{K}_{b}} base dissociation constant as:
Kb=[OH−]2[Sr2+][Sr(OH)2] Kb=[2x]2[x][0.1475−x] =6.5×10−3 \begin{aligned} & {{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{aligned}
Now, we will solve the value of x as:
x= 0.05346 M.
Now, we can write[OH−]\left[ O{{H}^{-}} \right] as:
[OH−]=2x=2×0.05346=0.1069M\left[ O{{H}^{-}} \right]=2x=2\times 0.05346=0.1069M
- As we know that in any aqueous solution, both ions that are [OH−]\left[ O{{H}^{-}} \right] and [H3O+]\left[ {{H}_{3}}{{O}^{+}} \right] are present and they must satisfy the following conditions:
[OH−][H3O+]=Kw [OH−][H3O+]=1.0×10−14 \begin{aligned} & \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{aligned} - As we have determined the value of [OH−]\left[ O{{H}^{-}} \right] as 0.1.69 M. Now, further we can solve as:
[H3O+]=1.0×10−14[OH−] [H3O+]=1.0×10−14[0.1069] [H3O+]=9.355−14M \begin{aligned} & \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{aligned} So, we get the value of [H3O+]\left[ {{H}_{3}}{{O}^{+}} \right] as 9.355−14M{{9.355}^{-14}}M .
- Now, we will calculate the value of pH by using the formula of pH as:
pH=−log⁡[H3O+] pH=−log⁡[3.955−14] pH=13.0290 \begin{aligned} & pH=-\log \left[ {{H}_{3}}{{O}^{+}} \right] \\\ & pH=-\log \left[ {{3.955}^{-14}} \right] \\\ & pH=13.0290 \\\ \end{aligned}
- We can calculate the value of pOH by using the formula:
pOH=−log⁡[OH−]pOH=-\log \left[ O{{H}^{-}} \right]
On calculating we get:
pOH=−log⁡[0.1069] pOH=0.9710 \begin{aligned} & pOH=-\log \left[ 0.1069 \right] \\\ & pOH=0.9710 \\\ \end{aligned} - Hence, we get the value of pH, and pOH as 13.0290 and 0.9710.

Note: - We should not get confused in the symbols Kb{{K}_{b}} and KB{{K}_{B}} . As Kb{{K}_{b}} is a base dissociation constant that basically measures how completely a base dissociates into its component ions in water.
- Whereas, KB{{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.