Question

What amount of sodium propanoate should be added to one liter of a aqueous solution containing 0.02 mole of propanoic acid [${{K}_{a}}=1.34\text{ x 1}{{\text{0}}^{-5}}\text{ at 2}{{\text{5}}^{\circ }}C$] to obtain a buffer solution at pH 4.75
(a)- $4.52\text{ x 1}{{\text{0}}^{-2}}\text{ M}$
(b)- $3.52\text{ x 1}{{\text{0}}^{-2}}\text{ M}$
(c)- $2.52\text{ x 1}{{\text{0}}^{-2}}\text{ M}$
(d)- $\text{1}\text{.7 x 1}{{\text{0}}^{-2}}\text{ M}$

Answer 1

Hint: The concentration of the salt can be calculated by the formula- $pH=p{{K}_{a}}+\log \frac{[Salt]}{[Acid]}$, where pH of the buffer solution is 4.75. $p{{K}_{a}}$ is the negative logarithm of the dissociation constant of the acid. The number of moles can be calculated by multiplying the concentration of the salt to the volume of the solution.


Complete step by step answer:
First, we have to calculate the concentration of the salt.
From the Henderson-Hasselbalch equation:
$pH=p{{K}_{a}}+\log \frac{[Salt]}{[Acid]}$
We can find the concentration of salt (sodium propanoate)
The number of moles of propanoic acid is given 0.02 and the volume of the solution is 1L.
So the concentration of the acid will be:
$[Acid]=\frac{0.02}{1}=0.02$
We have ${{K}_{a}}=1.34\text{ x 1}{{\text{0}}^{-5}}$, and we know that $p{{K}_{a}}$ is the negative logarithm of the dissociation constant of the acid (${{K}_{a}}$).
So, $p{{K}_{a}}=-\log {{K}_{a}}$
$p{{K}_{a}}=-\log (1.34\text{ x 1}{{\text{0}}^{-5}})$
$p{{K}_{a}}=4.870$
And given the pH of the buffer solution is 4.75
Now putting all these in the formula, we get
$pH=p{{K}_{a}}+\log \frac{[Salt]}{[Acid]}$
$4.75=4.870+\log \frac{[Salt]}{[0.02]}$
$\log \frac{[Salt]}{[0.02]}=-0.12$
$[Salt]=Anti\log (-0.12)\text{ x 0}\text{.02}$
$[Salt]=0.017\text{ mol/L}$
So, the concentration of the salt (sodium propanoate) is $0.017\text{ mol/L}$
And we know the volume of the solution is 1L
The number of moles can be calculated by multiplying the concentration of the salt to the volume of the solution.
$moles=conc\text{ x }volume$
$moles=0.017\text{ x 1 = 0}\text{.017 = 1}\text{.7 x 1}{{\text{0}}^{-2}}$
So, the moles required is $\text{1}\text{.7 x 1}{{\text{0}}^{-2}}\text{ M}$

Hence, the correct answer is an option (d)- $\text{1}\text{.7 x 1}{{\text{0}}^{-2}}\text{ M}$


Note: If we want to calculate the concentration of conjugate base, the formula used is $pH=p{{K}_{a}}+\log \frac{[Conjugate\text{ }base]}{[Acid]}$ . If the concentration of base is given then the formula is $pH=p{{K}_{a}}+\log \frac{[Salt]}{[Base]}$.
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