Answer to Question #148122 in General Chemistry for Chrishan

Question

The Ka for benzoic acid is 6.5 x 10-5
.
Calculate the pH of a 0.15 M benzoic acid
solution.

Answer 1

K_a= 6.5 x 10^-5

[C₆H₅COOH]= 0.15M

C₆H₅COOH:::::::: C₆H₅COO^- + H⁺

The equilibrium concentrations are

[C₆H₅COOH] = 0.15 - x

[C₆H₅COO^-] = x

[H⁺] = x

Now,

K_a= [C₆H₅COO^-][H⁺]/[C₆H₅COOH]

6.5x10^-5 = x²/0.15-x

Let's make an approximation 0.15-x is equal to 0.15 and then check for the validity of our approximation

x²/0.15 = 6.5x10^-5

x²= 4.33 x 10^-4

x= 0.021M

Let's check for the validity of our approximation

Percent ionization of benzoic acid= x/[C₆H₅COOH] x 100%

= 0.021/0.51 x 100%

= 14%

Our approximation is invalid, approximations of above 5% ionization are generally invalid, therefore we need to set up a quadratic equation.

6.5x10^-5 = x²/ 0.15-x

x²= 6.5x10^-5 (0.15-x)

x² = 9.75x10^-6-6.5x10^-5x

x² + 6.5x10^-5x-9.75x10^-6= 0

Using the formula

x= -b+- √b² - 4ac/2a

Where a= 1, b= 6.5 x10^-5, c= 9.75x10^-6

x= -6.5x10^-5 +- √(6.5x10^-5)² - (4)(1)(-9.75x10^-6)/2(1)

x= -6.5 x 10^-5 +- 6.28x 10^-3/2

x= 3.11 x 10^-3

Therefore, [H⁺] = 3.11 x 10^-3M

pH= -log[H⁺]

pH = -log[3.11x10^-3]

pH= 2.51