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ICE Princess25 [194]
3 years ago
12

Calculate the pH of a solution prepared by dissolving 0.150 mol of benzoic acid and 0.300 mol of sodium benzoate in water suffic

ient to yield 1.00 L of solution. The Ka of benzoic acid is 6.30 × 10-5.
Chemistry
1 answer:
jok3333 [9.3K]3 years ago
8 0

Answer : The pH of a solution is, 4.50

Explanation : Given,

K_a=6.30\times 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_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

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

pK_a=5-\log (6.30)

pK_a=4.20

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

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

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

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

pH=4.50

Thus, the pH of a solution is, 4.50

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