Question

riaz wants to prepare a buffer solution of ph 4.0 containing formic acid and sodium formate. he knows that pka for formic acid is 3.80. what volume of 0.10 m sodium formate solution should he add to 50 ml of 0.05 m formic acid to prepare the buffer solution?

Answer 1

Answer : The volume of sodium formate solution added = 39.63 ml

Solution : Given,

ph of buffer solution = 4

Dissociation constant of acid, $p_{ka}$ = 3.8

Molarity of sodium formate = 0.1 m

Molarity of formic acid = 0.05 m

Volume of formic acid = 50 ml

Formula used :

For acidic buffer,

$p_h=p_{ka}+log\frac{[salt]}{[acid]}$            ...............(1)

First we have to calculate the concentration of sodium formate (salt) and formic acid.

As we know,      $Molarity=\frac{\text { Total millimoles}}{\text{ Total volume}}$

Now the concentration of sodium formate [salt] is,

$[salt]=\frac{M_1\times V_1 }{V}$              ...........(2)

where,

$M_1$ = molarity of sodium formate

$V_1$ = volume of sodium fromate

V = Total volume = $V_1+V_2$

Now put all the values in above expression (2), we get

$[salt]=\frac{0.1\times V_1}{V_1+50}$         .............(3)

Similarly, for the formic acid concentration.

$[acid]=\frac{M_2\times V_2}{V}$             .............(4)

where,

$M_2$ = molarity of formic acid

$V_2$ = volume of formic acid

V = Total volume = $V_1+V_2$

Now put all the values in above expression (4), we get

$[acid]=\frac{0.05\times 50}{V_1+50}$          ............(5)

Now we have to calculate the volume of sodium formate solution by using above equation (1),

$4=3.8+log\frac{(\frac{0.1\times V_1}{V_1+50})}{(\frac{0.05\times 50}{V_1+50})}$

By rearranging the terms, we get

$V_1$ = 39.63 ml

Therefore, the volume of sodium formate solution added = 39.63 ml