Question

1) What is the pH of a 1.5x10^-8 M solution of HCl? 2) What is the...

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.

Homework Answers

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

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