When 14.7 mL of aqueous HBr (a strong acid) was added to water, 0.482 L of a solution with a pH of 4.23 was produced. What was the molarity of the original HBr solution?

Question

Quizplus · Accepted Answer

The correct answer is A because the molarity of the original HBr solution can be calculated using the formula: M1V1 = M2V2, where M1 is the molarity of the original HBr solution, V1 is the volume of the original HBr solution, M2 is the molarity of the diluted solution, and V2 is the volume of the diluted solution. Substituting the given values into the formula, we get: M1 * 14.7 mL = 0.482 L * 10^-4 M. Solving for M1, we get: M1 = 0.482 L * 10^-4 M / 14.7 mL = 1.9 × 10^-3 M.

When 147 ML of Aqueous HBr (A Strong Acid) Was Added