Question

To a $50mL$ of $0.05M$ formic acid how much volume of $0.10M$ sodium formate must be added to get a buffer solution of $pH = 4.0$? ($p{K_a}$ of acid is $3.7$)
A: $40mL$
B: $4mL$
C: $50mL$
D: $100mL$

Answer 1

A buffer solution is an aqueous solution that is made up of weak acid and its conjugate base or weak base and its conjugate acid. The $pH$ of buffer solution changes very little when a small amount of strong acid or base is added in this solution.
Formula used: $p{K_a} = - \log \left( {{K_a}} \right)$
$pH = - \log \left( {{K_a}} \right) + \log \left( {\dfrac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}} \right) = p{K_a} + \log \left( {\dfrac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}} \right)$
Concentration $= \dfrac{{{\text{total millimole}}}}{{{\text{total volume}}}}$
Total milli mole$= molarity \times volume$
Where, $\left[ {salt} \right]$ is concentration of salt formed and $\left[ {acid} \right]$ is concentration of the acid.

Complete step by step answer:
A buffer solution is an aqueous solution that is made up of weak acid and its conjugate base or weak base and its conjugate acid. Conjugate base is a compound that is formed when an acid loses its hydrogen ion or we can say that when an acid loses its hydrogen ion conjugate base is formed. Similarly conjugate acid is formed when a base loses its $O{H^ - }$ ion. The $pH$ of buffer solution changes very little when a small amount of strong acid or base is added in this solution. $pH$ is the measure of concentration of hydrogen ions in a solution.
In this question we have been given an acidic buffer. For this buffer, acid is formic acid and salt of this acid is sodium formate. Chemical formula of formic acid is $HCOOH$ and the chemical formula of sodium formate is $HCOONa$.
Total volume of the solution will be the sum of the volume of formic acid and sodium formate. We have to find the volume of sodium formate. So, let the volume of sodium formate added be $V$.
We have given $50mL$ of $0.05M$ formic acid. Therefore molarity of formic acid is $0.05M$ and volume of formic acid is $50mL$.
This means total volume of solution is $\left( {V + 50} \right)L$.
Formula to calculate millimoles is:
milli mole$= molarity \times volume$
millimoles of formic acid will be,
millimole$= 0.05 \times 50 = 2.50$ (it is given that molarity of formic acid is $0.05M$ and volume of formic acid is $50mL$)
given molarity of sodium formate is $0.10M$ and we have supposed the volume of sodium formate $V$. Substituting these values in the formula of millimoles, we get,
millimole$= 0.10 \times V = 0.10V$
Now, we have millimoles of both acid and the salt formed and the total volume of solution is $\left( {V + 50} \right)L$ (explained above). This means we can calculate the concentration of acid and salt. Formula to calculate concentration is:
Concentration $= \dfrac{{{\text{total millimole}}}}{{{\text{total volume}}}}$
Milli Mole of formic acid is $2.50$ and total volume is $\left( {V + 50} \right)L$ (calculated above). Substituting these values we get:
Concentration of formic acid$= \dfrac{{2.50}}{{V + 50}}$
Millimoles of sodium formate is $0.10V$ and total volume is $\left( {V + 50} \right)L$ (calculated above). Substituting these values we get:
Concentration of sodium formate$= \dfrac{{0.10V}}{{V + 50}}$
Formula to calculate $pH$ of acidic buffer is:
$pH = p{K_a} + \log \left( {\dfrac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}} \right)$
$pH$ of given buffer solution is $4.0$ and $p{K_a}$ of acid is $3.7$. Concentration of salt and acid is calculated above. Substituting these values in the formula we get:
$4 = 3.7 + \log \left( {\dfrac{{\dfrac{{0.1V}}{{V + 50}}}}{{\dfrac{{2.5}}{{V + 50}}}}} \right)$
This equation can be written as:
$4 = 3.7 + \log \left( {\dfrac{{0.1V}}{{2.5}}} \right)$
Solving this we get,
$V = 49.88 \approx 50L$
Therefore volume of sodium formate is $50L$.
Hence the correct option is C.

Note:
Blood is also a buffer solution. $pH$ of blood is $7.4$. Buffer system that is present in our blood is ${H_2}C{O_3} + NaHC{O_3}$.
Buffer capacity of a solution is maximum when the concentration of the weak acid and its salt are equal.