What volume of $0.10 M$ sodium formate solution should be added to $50 m L$ of $0.05 M$ formic acid to produce a buffer solution of pH $4.0$ ? $p K_{a}$ for formic acid is $3.80$

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Solution

Let $x$ mL of $0.10 M$ sodium formate be added
No. of moles in $x m L$ of $0.10 M$ sodium formate $= \frac{0.10}{1000} \times x$
No. of moles in $50 m L$ of $0.05 M$ formic acid $= \frac{0.05}{1000} \times 50$
$\frac{[S o d . f o r m a t e]}{[F o r m i c a c i d]} = \frac{0.10 x}{2.5} = 0.04 x$
Applying the equation
$p H = log \frac{[S a l t]}{[A c i d]} + p K_{a}$
$4.0 = log 0.04 x + 3.8$
or $x = 39.6 m L$

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What volume of 0.10M sodium formate solution should be added to 50mL of 0.05M formic acid to produce a buffer solution of pH 4.0? pKa for formic acid is 3.80

Let x mL of 0.10M sodium formate be addedNo. of moles in xmL of 0.10M sodium formate=0.101000×xNo. of moles in 50mL of 0.05M formic acid=0.051000×50[Sod.formate][Formic acid]=0.10x2.5=0.04xApplying the equationpH=log[Salt][Acid]+pKa4.0=log0.04x+3.8or x=39.6mL

What volume of $0.10 M$ sodium formate solution should be added to $50 m L$ of $0.05 M$ formic acid to produce a buffer solution of pH $4.0$ ? $p K_{a}$ for formic acid is $3.80$