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GarryVolchara [31]
3 years ago
13

Calculate the pH of a solution prepared by dissolving 0.15 mol benzoic acid (C7H5O2H) and 0.30 mol of sodium benzoate (Na C7H5O2

) in water sufficient to yield 1.00 L of solution. The Ka of benzoic acid is 6.50 x 10-5
Chemistry
1 answer:
asambeis [7]3 years ago
8 0

Answer : The pH of a solution is, 4.5

Explanation : Given,

Moles of benzoic acid = 0.15 mol

Moles of sodium benzoate = 0.30 mol

Volume of solution = 1.00 L

The dissociation constant for benzoic acid = K_a=6.5\times 10^{-5}

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 (6.5\times 10^{-5})

pK_a=5-\log (6.5)

pK_a=4.2

Now we have to calculate the pH of a solution.

Using Henderson Hesselbach equation :

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

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

pH=4.2+\log [\frac{(\frac{0.30}{1.00L})}{(\frac{0.15}{1.00L})}]

pH=4.5

Therefore, the pH of a solution is, 4.5

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The given question is incomplete. the complete question is:

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