Question

What volume of 0.1M Ch3cooh is to be added to 50ml of 0.2 M Na acetate to prepare a buffer solution of pH 4.91?

Answer 1

Answer:

55 ml of volume of 0.1M $\mathrm{CH}_{3} \mathrm{COOH}$ is to be added to 50ml of 0.2 M Na acetate to prepare a buffer solution of pH 4.91.

To find:

The volume of “0.1 mole of acetic acid”.

Solution:

By using the formula of the pH,

$p H=p K_{a}+\log \frac{[\text {Salt}]}{[A c i d]}$

$\log \frac{[\text {Salt}]}{[\text {Acid}]}=p H-p K_{a}$

pH of acetic acid=4.74

=4.91-4.74=0.17

$\log \frac{[\text {Salt}]}{[\text {Acid}]}=0.17$

$\log \frac{[\text { Salt }]}{[A c i d]}=1.82$

Number of moles of sodium acetate in the solution $=\frac{50 \times 0.2}{1000}$

=0.01 mol

Suppose volume of 0.2 M acetic acid added =V ml

Number of moles of acetic acid $=\frac{V \times 0.1}{1000}$

$\frac{0.01}{\frac{V \times 0.1}{1000}}=1.82$

V=55 ml

Then, 55 ml of 0.1 Mole of “acetic acid is needed” to add in the given condition.

Answer 2

Answer:

71 mL of acetic acid should be added to sodium acetate to prepare buffer solution of pH = 4.91

Explanation:

To find the volume of acetic acid.

Let us assume volume of acetic acid as V ml.

From the given statement we have,

Concentration of acetic acid = 0.1 M.

Therefore,

Number of mole in acetic acid = 0.1 V mill mole.  

Concentration of sodium acetate = 0.2 M

Volume of sodium acetate solution= 50 ml  

Therefore,

Number of mole of sodium acetate = 0.2 × 50 = 10 millimole

Now,

Concentration of acetic acid, [acid] = 0.1 V/(V + 50) M  

Concentration of sodium acetate, [salt] = 10/(V + 50) M

Hence,

pH = pKa + log[salt]/[acid]  

By substituting the values

4.91 = 4.76 + log[10/0.1V]  

4.91 - 4.76 = log(100/V)  

0.15 = log(100/V)

$10^{0.15}=\frac{100}{V}$

$V=\frac{100}{10^{0.15}}$ ≈ 71mL

Explanation: