Question

What volume of 1.10 M sodium formate solution should be to 50 mL of 0.05 M formic acid to produce a buffer solution of pH 4.0? (pK_a for formic acid is 3.80).

• A. 3.96 mL
• B. 25 mL
• C. 39.6 mL
• D. 100 mL

Answer 1

The correct option is C

39.6 mL

Let $x$ mL of $0.10$ M sodium formate is added.
Number of moles in $xmL$ of $0.10M$ sodium formate $=\frac{1.10}{1000}\times x$

Number of moles in $50mL\times mL$ of $0.05M$ formic acidformate $=\frac{0.05}{1000}\times 50$

$\therefore \frac{\left[\text{soditumformate}\right]}{\left[\text{Formicacid}\right]}=\frac{\frac{0.10\times x}{1000}}{\frac{0.05\times 50}{1000}}=0.04x$

From Henderson's equation,

$pH={pK}_a+\log \frac{\left[\text{salt}\right]}{\left[\text{acid}\right]}$
$4.0=3.80+\log 0.04x$
$\log 0.04x=0.2$
$\therefore x=39.6mL$