What is the pH of a 0.05M solution of formic acid? $({{K}_{a}}=1.8\times {{10}^{-4}})$
A) 5.0
B) 1.8
C) 2.52
D) 0.5

Hint: Calculation of pH on the basis of dissociation of formic acid with the help of the formula for ${{H}^{+}}$ on the basis of definition of pH given by $pH = -\log \left[ {{H}^{+}} \right]$ will lead you to the required answer.


Complete Solution :
We have studied in our lower classes of physical chemistry about the dissociation constants and also about initial and final concentration change.
We know that pH is defined as the negative logarithm of the hydrogen ion concentration present in the given solution which decides the acidity and basicity of the solution.
Let us now see the calculation of pH on the basis of concentration change at different time intervals that is initial and final.
Now, the dissociation of formic acid is as shown below which dissociates into ${{H}^{+}}$ and an anion as



Now, the dissociation constant can be written by ignoring (1-x) because the value of dissociation constant is very small.
Therefore,${{K}_{a}}=0.05{{x}^{2}}$
By substituting the values we get,
$1.8\times {{10}^{-4}}=0.05{{x}^{2}}$
\[\Rightarrow x=\sqrt{\dfrac{1.8\times {{10}^{-4}}}{0.05}}\]
Thus, by solving we get$x=5.47\times {{10}^{-2}}$

- Now, we need to calculate pH and thus the concentration of $[{{H}^{+}}]=0.05x=0.05\times 5.47\times {{10}^{-2}}$
\[\Rightarrow [{{H}^{+}}]=0.3\times {{10}^{-2}}\]
Now, $pH=-\log \left[ {{H}^{+}} \right]=-\log \left( 0.3\times {{10}^{-2}} \right)$
By splitting the log values and simplifying the signs we get
\[\Rightarrow pH=-\log 0.3 + 2\log 10\]
\[\Rightarrow pH=0.52 + 2 = 2.52\]
Therefore, the pH of 0.05 M solution of formic acid is 2.52
So, the correct answer is “Option C”.

Note: pH calculation sometimes can also be written as $pH = -\log \left[ {{H}_{3}}{{O}^{+}} \right]$. But both the formulas are one and the same for the calculation of negative logarithm of hydrogen ion concentration and do not get wrong answers when the question is based on this formula.