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The ka of benzoic acid is 6.30 ⋅ 10-5. The ph of a buffer prepared by combining 50.0 ml of 1.00 m potassium benzoate and 50.0 ml of 1.00 m benzoic acid is ________.

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Answer;

pH = 4.20

Explanation;

pKa = -log(6.30 × 10^-5)

pKa = 4.20  

Moles of Benzoic acid = volume × molarity

                                      = 0.050L × 1.00M

                                      = 0.050moles benzoic acid  

Moles of the salt = 0.050L ×  1.00M

                           = 0.050 moles salt  

Therefore;

0.050mols / 0.1 L = 0.50M  

0.050mols / 0.1 L = 0.50M  

Thus;

pH = 4.20 + log(0.50/0.50)

pH = 4.20

pstnonsonjoku

From the calculations, the pH of the solution is 4.20

What is pH?

The pH is defined as the degree of acidity or alkalinity of a solution. We have to find the pH of the buffer using the Henderson–Hasselbalch equation.

Since, Ka = 6.30 × 10^-5

then pKa = -log(6.30 × 10^-5)

pKa = 4.20  

Number of moles of Benzoic acid = volume × molarity

= 0.050L × 1.00M = 0.050moles benzoic acid  

Number of moles of benzoate = 0.050L ×  1.00M = 0.050 moles

Total volume of solution = 50 mL + 50 mL = 100 mL or 0.1 L

Molarity of acid

0.050mols / 0.1 L = 0.50M  

Molarity of conjugate base

0.050mols / 0.1 L = 0.50M  

Hence;

pH = 4.20 + log(0.50/0.50)

pH = 4.20

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