# calculate the ph of a solution prepared by dissolving 0.150 mol of benzoic acid and 0.300 mol of sodium benzoate in water sufficient to yield.1.00l of solution

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Answer : The pH of a solution is, 4.50

Explanation : Given,

K_a=6.30\times 10^{-5}K

a

=6.30×10

−5

Concentration of benzoic acid (Acid) = 0.150 M

Concentration of sodium benzoate (salt) = 0.300 M

First we have to calculate the value of pK_apK

a

.

The expression used for the calculation of pK_apK

a

is,

pK_a=-\log (K_a)pK

a

=−log(K

a

)

Now put the value of K_aK

a

in this expression, we get:

pK_a=-\log (6.30\times 10^{-5})pK

a

=−log(6.30×10

−5

)

pK_a=5-\log (6.30)pK

a

=5−log(6.30)

pK_a=4.20pK

a

=4.20

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}pH=pK

a

+log

[Acid]

[Salt]

Now put all the given values in this expression, we get:

pH=4.20+\log (\frac{0.300}{0.150})pH=4.20+log(

0.150

0.300

)

pH=4.50pH=4.50

Thus, the pH of a solution is, 4.50

Hope it helps ☺️!!!