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Question: What value of \(0.1M\) HCOONa should be added to \(0.05M\) \(50ML\) HCOOH to make a buffer of \(pH =...

What value of 0.1M0.1M HCOONa should be added to 0.05M0.05M 50ML50ML HCOOH to make a buffer of pH=4pH = 4 if pKa=3.7 pKa = 3.7

Explanation

Solution

In the given question we have to make a buffer solution of the given salt (sodium acetate) and acid (acetic acid) of pH 4. It can be done by the Henderson equation. By using the given values one can easily determine the buffer solution.

Complete step by step solution:
Let xMLxML of 0.1M0.1M HCOONa is added.
Number of moles in xmLx\,mL of 0.1M0.1M HCOONa =0.11000×x = \dfrac{{0.1}}{{1000}} \times x
Number of moles in 50mL50\,mL of 0.05M0.05M formic acid formate =0.051000×50 = \dfrac{{0.05}}{{1000}} \times 50
Therefore [HCOONA][HCOOH]=0.11000×x0.051000×50=0.04x\dfrac{{[HCOONA]}}{{[HCOOH]}} = \dfrac{{\dfrac{{0.1}}{{1000}} \times x}}{{\dfrac{{0.05}}{{1000}} \times 50}} = 0.04x
From Henderson–Hasselbalch equation we can calculate the pH of a solution containing acid and one of its salts, that is, of a buffer solution.
From Henderson equation,
pH=pKa+log[HCOONA][HCOOH]pH = pKa + \log \dfrac{{[HCOONA]}}{{[HCOOH]}}
By putting values in equation we get,
4=3.7+log0.04x4 = 3.7 + \log 0.04x
log0.04x=0.3\log 0.04x = 0.3
x=38.6x = 38.6

Formula used:
Henderson equationpH=pKa+log[salt][acid]pH = pKa + \log \dfrac{{[salt]}}{{[acid]}}
Where,
pHpH=acidity of a buffer solution
pKapKa=negative logarithm of Ka
KaKa=acid dissociation constant
[acid]=concentration of an acid
[salt]=concentration of conjugate base

Additional information:
Buffers are of two types which include acidic and alkaline buffer solutions. Acidic buffers have a pH below 7 and solution contains a weak acid and one of its salts. For example, a mixture of acetic acid and sodium acetate acts as a buffer solution with a pH of about 4.
Alkaline buffers have a pH above 7 and solution contains a weak base and one of its salts. For example, a mixture of ammonium chloride and ammonium hydroxide acts as a buffer solution with a pH of about 9.

Note:
The Henderson-Hasselbalch equation is used for the estimation of the pH of a buffer solution and also to find the equilibrium pH in an acid-base reaction. In the reaction equilibrium between the weak acid and its conjugate base makes the solution to oppose changes to pH when some amounts of strong acid or base are added to it.