Answer to Question #21499 in Physical Chemistry for Celebilover

The formula for K_a is
K_a = [H⁺][B^-]/[HB]

where
[H⁺] = concentration of H⁺ ions
[B^-] = concentration of conjugate base ions
[HB] = concentration of undissociated acid molecules
for a reaction HB → H⁺ + B^-

Let x represent the concentration of H⁺ that dissociates from HB, then [HB] = C - x where C is the initial concentration.

Enter these values into the K_a equation

K_a = x · x / (C -x)
K_a = x^²/(C - x)
(C - x)K_a = x^²
x^² = CK_a - xK_a
x² + K_ax - CK_a = 0

Solve for x using the quadratic equation

x = [-b ± (b^² - 4ac)^½]/2a

x = [-K_a + (K_a^² + 4CK_a)^½]/2

**Note** Technically, there are two solutions for x. Since x represents a concentration of ions in solution, the value for x cannot be negative.

Enter values for K_a and C

K_a = 1x10^-7
C = 0.1 M

x = {-1x10^-7 + [(1x10^-7)^² + 4(0.1)(1x10^-7)]^½}/2
x = 0.0001

[H⁺] = 0.0001

If [H⁺] = 0.0001, [OH^-] = 1x10^-10

c) [H⁺] = 0.0001, [OH^-] = 1x10^-10