Question

Calculate change in pH upon ten-fold dilution of the following solutions:
(a) 0.1 HCl
(b) 0.1M acetic acid
(c) 0.1M $N{{H}_{4}}Cl$
$OH=1.8\times {{10}^{-5}},{{K}_{b}}\left( N{{H}_{3}} \right)=1.8\times {{10}^{-5}}$

Answer 1

Hint:To calculate the pH value we should know the concentration of the ions present in the solution and the equation used for calculating pH is as follows:
$pH=-\log \left[ {{H}^{+}} \right]$, where ${{H}^{+}}$ is the concentration of the ions.

Complete step-by-step answer:In the question given, we have to find the pH of various solutions and the initial concentration of the ions present in the solution at zero time is given.
Let us solve the question one by one:
(a) Here the pH of HCl should be calculated, we know that HCl is a strong acid and the acid will get completely dissociated.
The reaction is as follows: $HCl\rightleftharpoons {{H}^{+}}+C{{l}^{-}}$
The concentration is given as 0.1M and let us find the pH of HCl.
$\left[ {{H}^{+}} \right]=0.1M={{10}^{-1}}M$
$\Rightarrow pH=-\log \left[ {{H}^{+}} \right]$
$\Rightarrow pH=-\log \left[ {{10}^{-1}} \right]=1$
When the solution is diluted to ten times then the concentration should be divided by 10.
Then we get the concentration as 0.01M.And now find the pH with this value of concentration.
$pH=-\log \left[ {{H}^{+}} \right]=-\log \left[ {{10}^{-2}} \right]$
$\Rightarrow pH=2$
Hence the pH values with change in concentration from 1 to 2.
(b) Now let us see the dissociation reaction of acetic acid.We know that the acetic acid is a weak acid so it does not dissociate completely in the solution.
The concentration of acetic acid given is 0.1M. Now let us plot the change of the concentration happening in the reactants and products.

The equation is-	$C{{H}_{3}}COOH\rightleftharpoons $	$C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}$
Change conc.	$\left( 0.1-x \right)$	$x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x$

Here $\left( 0.1-x \right)\approx 0.1$M
Now let us find the value x, i.e. the amount of products formed.
$\dfrac{x\times x}{0.1-x}=\dfrac{{{x}^{2}}}{0.1}=1.8\times {{10}^{-5}}$
$\Rightarrow {{x}^{2}}=1.8\times {{10}^{-5}}\times 0.1=1.8\times {{10}^{-6}}$
$\Rightarrow x=\sqrt{1.8\times {{10}^{-6}}}$
$\Rightarrow x=1.34\times {{10}^{-3}}$
$pH=-\log x$
$pH=-\log \left( 1.34\times {{10}^{-3}} \right)=2.87$
When the solution is diluted by 10 times, then the pH will change. The concentration is changing from 0.1M to 0.01M.
$\dfrac{{{x}^{2}}}{0,01}=1.8\times {{10}^{-5}}$
$\Rightarrow {{x}^{2}}=1.8\times {{10}^{-5}}\times 0.01=1.8\times {{10}^{-7}}$
$\Rightarrow x=4.24\times {{10}^{-4}}M$
$pH=-\log x=-\log \left( 4.24\times {{10}^{-4}} \right)=3.37$
(c) The$N{{H}_{4}}Cl$, is a salt which is formed by weak base and strong acid.

$N{{H}_{4}}^{+}+{{H}_{2}}O\rightleftharpoons$	$N{{H}_{4}}OH+{{H}^{+}}$
$\left( 0.1-h \right)$	$h \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,h$

Now let us find the value of h first i.e. the change in the concentration.
$\dfrac{{{h}^{2}}}{0.1}={{K}_{h}}$
$\Rightarrow {{h}^{2}}={{K}_{h}}\times 0.1$
$\Rightarrow {{K}_{h}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}$
$\Rightarrow {{K}_{w}}={{10}^{-14}}$ and ${{K}_{b}}=1.8\times {{10}^{-5}}$
Substitute the values in the equation and we get,
${{K}_{h}}=\dfrac{{{10}^{-14}}}{1.8\times {{10}^{-5}}}=5.5\times {{10}^{-10}}$
\[\Rightarrow {{h}^{2}}=0.1\times 5.5\times {{10}^{-6}}=5.5\times {{10}^{-7}}\]
$\Rightarrow h=7.41\times {{10}^{-6}}=\left[ {{H}^{+}} \right]$
$pH=-\log \left( 7.41\times {{10}^{-6}} \right)=5.130$
When the solution is diluted by 10 times the concentration gets reduced by 10 times and the concentration will be 0.01M.
${{h}^{2}}=0.01\times {{K}_{h}}$
$\Rightarrow {{h}^{2}}=0.01\times 5.5\times {{10}^{-10}}=5.5\times {{10}^{-12}}$
$\Rightarrow h=2.3452\times {{10}^{-6}}$
$pH=-\log \left( 2.3452\times {{10}^{-6}} \right)=5.629$
The pH will change from 5.130 to 5.629 on dilution.

Note:Always write the reaction associated to trace the change in the concentration terms.
Always take care while doing the calculations with variables and minimize the calculations to avoid the confusions. First carry out the mathematical operations like division and multiplication regarding the terms whose logarithmic value has to be taken. Simplify the calculations as much as possible and then take the logarithmic value of the number for accurate answers.