Question

What amount of sodium propanoate should be added to one litre of an aqueous solution containing $0.02mole$ of propanoic acid $(K_a=1.34\times {10}^{-5}$ at ${25}^{\circ }C$) to obtain a buffer solution of $pH4.75$.

• A. $4.52\times {10}^{-2}M$
• B. $3.52\times {10}^{-2}M$
• C. $2.52\times {10}^{-2}M$
• D. $1.7\times {10}^{-2}M$

Answer 1

Answer: (A)

$4.52\times {10}^{-2}M$

$pH=p{K}_a+log\frac{\left[Salt\right]}{\left[Acid\right]}$ Given, $pH=4.75$

$4.75=4.870+log\frac{\left(salt\right)}{\left(0.02\right)}$ $p{K}_a=-log{K}_a=4.870$

$log\frac{\left(salt\right)}{\left(0.02\right)}=-0.12$

(salt) = Antilog$(-0.12)\times 0.02$

(salt) = $0.017$ mol/L

No. of moles required $=conc\times vol=0.017moles$