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What is the pH of a 1.0 M solution of acetic acid? To what volume of 1 L of this solution be diluted so that the pH of the resulting solution will be twice the original value? Given, ${ K }_{ a }=1.8\times { 10 }^{ -5 }$.

Answer Verified Verified
Hint: The pH of a dilute solution is equal to the negative logarithm of the hydronium ion concentration. If the solution is concentrated then it will be equal to the negative logarithm of the activity of the hydronium ions.

Complete answer:
To find, the pH of the solution, consider the reaction:
$ { CH }_{ 3 }COOH(aq)\rightleftharpoons { H }^{ + }(aq)+{ CH }_{ 3 }{ COO }^{ - }(aq)$
$ { K }_{ a }=\cfrac { \left[ { H }^{ + } \right] \left[ { CH }_{ 3 }{ COO }^{ - } \right] }{ \left[ { CH }_{ 3 }COOH \right] } $
Let the concentration of $ { H }^{ + }$ be x. Hence $ \left[ { H }^{ + } \right]$=$ \left[ { CH }_{ 3 }{ COO }^{ - } \right]$=x
Hence the equation will become:
$ { K }_{ a }=\cfrac { { x }^{ 2 } }{ 1-x } $
Since acetic acid is a weak acid, we will ignore the value of x with respect to 1:
$ \Rightarrow { K }_{ a }=\cfrac { { x }^{ 2 } }{ 1 } $
$ { x }^{ 2 }=\sqrt { 1.8\times { 10 }^{ -5 } } =4.243\times { 10 }^{ -3 }$
Therefore the pH will be:
$ pH=-log(4.243\times { 10 }^{ -3 })=2.37$
Now we need to dilute this 1 L of solution such that the pH of the final solution will be 4.74 (twice the original). Hence the concentration of the $ { H }^{ + }$ will be:
$pH=4.74$
$\Rightarrow -\log { \left[ { H }^{ + } \right] } =4.74$
$\Rightarrow \left[ { H }^{ + } \right] ={ 10 }^{ -4.74 }=1.820\times { 10 }^{ -5 }$
Now, we will apply the following equation:
${ M }_{ 1 }{ V }_{ 1 }={ M }_{ 2 }{ V }_{ 2 }$
$\Rightarrow (4.243\times { 10 }^{ -3 })\times 1\quad L=(1.820\times { 10 }^{ -5 }){ V }_{ 2 }$
Hence, ${ V }_{ 2 }=\cfrac { (4.243\times { 10 }^{ -3 })\times 1\quad L }{ (1.820\times { 10 }^{ -5 }) } =233.13\quad L$
Hence the pH of 1.0 M Acetic acid solution is 2.37. The volume to which 1 L of this solution should be diluted so that the pH of the resulting solution will be twice the original value is 233.13 L.

Note: x could only be neglected with respect to 1 if its value is less than 5% i.e.0.05. If its value is greater than or equal to 0.05 then it could not be neglected with respect to 1 and the equation will become a quadratic equation.