Question

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 ________.

Answer 1

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