Question

0.375 g of a monoprotic acid (mm = 245 g/mol) is dissolved in water to produce 25.0 mL of a solution with pH = 3.28. Determine the ionization constant of the acid.

Answer 1

Answer: 4.5 × 10^-6.

Explanation: 0.375 grams of a monoprotic acid is dissolved in water to produce 25.0 milliliters of a solution with pH = 3.28. this would give you 4.5 times 10 to the power of negative six.