What amount of sodium propanoate should be added to 1 liter of an aqueous solution containing 0.02 mol of propanoic acid (K a =1.0 × 10-5. at .25° C ) to obtain a buffer solution of pH 6 ?

Question

What amount of sodium propanoate should be added to 1 liter of an aqueous solution containing 0.02 mol of propanoic acid (K a =1.0 × 10-5. at .25° C ) to obtain a buffer solution of pH 6 ? (A) 0.1 M (B) 0.2 M (C) 0.3 M (D) 1.3 M

Tardigrade Admin · Accepted Answer

Using the expression pH =p K a + log ([ text Salt ]/[ text Acid ]) We get, 6=- log (1.0 × 10-5)+ log ([ text Salt ]/[0.02 M ]) Which gives 6=5+ log ([ text Salt ]/[0.02 M ]) or ([ Salt ]/[0.02 M ])=10 or [ textSalt ]=0.2 M

Q. What amount of sodium propanoate should be added to $1$ liter of an aqueous solution containing $0.02 m o l$ of propanoic acid $(K_{a} = 1.0 \times 1 0^{- 5}$ at $2 5^{\circ} C)$ to obtain a buffer solution of $p H 6$ ?

0.1 M

0.2 M

0.3 M

1.3 M

Using the expression

$p H = p K_{a} + lo g \frac{[ Salt ]}{[ Acid ]}$

We get, $6 = - lo g (1.0 \times 1 0^{- 5}) + lo g \frac{[ Salt ]}{[ 0.02 M ]}$

Which gives $6 = 5 + lo g \frac{[ Salt ]}{[ 0.02 M ]}$

or $\frac{[ S a lt ]}{[ 0.02 M ]} = 10$ or $[Salt] = 0.2 M$

Q. What amount of sodium propanoate should be added to 1 liter of an aqueous solution containing 0.02mol of propanoic acid (Ka​=1.0×10−5 at 25∘C) to obtain a buffer solution of pH6 ?