# The concentrations of acetic acid and sodium acetate is a buffer solution of total moalrity 0.39 M and having a pH of 4.4 if the ${K}_{a}$ of acetic acid is  is $1.83×{10}^{-5}$ is

1. A

0.26;  0.13

2. B

0.13; 0.26

3. C

0.30; 0.09

4. D

0.18; 0.21

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### Solution:

${K}_{a}=10-4.2=6.31×{10}^{-5}$,$log\frac{\left[Salt\right]}{\left[Acid\right]}=pH-p{K}_{a}$

$=pH+log{K}_{a}$=$4.4+log\left(1.83×{10}^{-5}\right)$= 4.4 – 4.7 = - 0.3

$\frac{\left[Salt\right]}{\left[Acid\right]}={10}^{-0.3}=0.5$  $1+\frac{\left[Salt\right]}{\left[Acid\right]}=1+0.5$

$\frac{\left[Acid\right]+\left[Salt\right]}{\left[Acid\right]}=1.5$ , $\frac{total\text{\hspace{0.17em}\hspace{0.17em}}molarity}{\left[Acid\right]}=1.5$$\frac{0.39}{\left[Acid\right]}=1.5$

$\left[Acid\right]=\frac{0.39}{1.5}=0.26$  [Salt] = 0.39 – 0.26 = 0.13

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