Calculate the pH of a solution made by mixing equal volumes of a solution of HCl with a pH of 1.52 and a solution of HNO₃ with a pH of 2.44. (Assume the volumes are additive.) A) 3.97 B) 1.98 C) 2.44 D) 1.77 E) 1.47

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First, let's calculate the concentration of H+ ions in each solution using the pH values: [H+] of HCl solution = 10^(-pH) = 10^(-1.52) = 3.52 x 10^(-2) M [H+] of HNO₃ solution = 10^(-pH) = 10^(-2.44) = 3.92 x 10^(-3) M When the two solutions are mixed, the volume of the resulting solution is double the volume of each original solution. Therefore, the concentration of H+ ions in the mixed solution can be calculated using the following equation: (H+ ion concentration of HCl solution x volume of HCl solution + H+ ion concentration of HNO₃ solution x volume of HNO₃ solution) / total volume of mixed solution Plugging in the values, we get: (H+ ion concentration of HCl solution x 0.5 + H+ ion concentration of HNO₃ solution x 0.5) / 1 = (3.52 x 10^(-2) M x 0.5 + 3.92 x 10^(-3) M x 0.5) / 1 = 1.95 x 10^(-2) M Finally, we can calculate the pH of the mixed solution using the formula: pH = -log[H+] Plugging in the value of [H+], we get: pH = -log(1.95 x 10^(-2)) = 1.77 Therefore, the answer is D.

Calculate the PH of a Solution Made by Mixing Equal