Answer to Question #179394 in General Chemistry for Kyle

Question

An erlenmeyer flask contains 50.0 mL of an aqueous solution of 0.425 M acetic acid (CH3COOH). Given that acetic acid has a Ka = 1.75 x10^(-5), answer the questions that follow.

(A) Calculate the pH of the resulting solution when the original solution is diluted to 250.0 mL via addition of distilled water. Write answer in THREE SIGNIFICANT FIGURES and complete solutions.

Answer 1

Given data-

Ka of acetic acid=1.75×10^-5

Concentration of acetic acid= 0.425 M

Given equation -

CH3COOH=CH3COO^- + H⁺

Calculate H⁺ using Ka equation -

Ka =[H⁺]² / acid

1.75×10^-5=[H⁺]² /. 425

[H⁺]² =1.75×10^-5 ×. 425=7.4375×10^-6

[H⁺] = 2.72×10^-3 M

We know,

P^H =-log[H⁺] = -log[2.72×10^-3]

P^H =2.56

Hence the P^H of the acetic acid is 2.56