Question

1) What is the pH of a 1.5x10^-8 M solution of HCl?

2) What is the pH when [HA-] = [A^2-] for a weak diprotic acid?

3) The formula for Malonic acid is HO2CCH2CO2H and it is a weak acid (K1 = 1.42x10^-3 , K2 = 2.01x10^-6). What is the pH of a 0.25M solution of HO2CCH2CO2Na?

4) Leucine is a diprotic amino acid (K1= 4.677 x 10^-3 , K2= 1.820 x 10^-10). Determine the pH if 11.0 mL of 1M NaOH are added to 1L of 0.01M of the acidic form of leucine.

Answer 1

1.

HCl (aq.) -----------> H⁺ (aq.) + Cl^- (aq.)

1.5 * 10^-8               1.5 * 10^-8  

H₂O (l) = H⁺ (aq.) + OH^- (aq.)

x M          x                  x    M

Total [H⁺] = x + (1.5 * 10^-8) M

Ionic product of water,

[H⁺][OH^-] = 1.0 * 10^-14

[ x + 1.5 * 10^-8] * [ x ] = 10^-14

x2 + 1.5 * 10^-8 x - 10^-14 = 0

Using quadratic equation,

x = [ - (1.5 * 10^-8) +/- sqrt. ((1.5*10^-8)² - (4 * 1 * (-10^-14)))] / ( 2 * 1)

x = [OH^-] = 9.27 * 10^-8 M

Now,

pOH = - Log[OH^-] = - Log(9.27 * 10^-8) = 7.03

Therefore,

pH + pOH = 14

pH = 14 - 7.03

pH = 6.97