Question

What is the pH of a 1.8 M solution of HClO4?

Assuming complete dissociation, what is the pH of a 3.06 mg/L Ba(OH)2 solution?

Enough of a monoprotic acid is dissolved in water to produce a 0.0112 M solution. The pH of the resulting solution is 2.65. Calculate the Ka for the acid.

Answer 1

(1) pH = -log [H+]

1.8 M HClO4 will release 1.8 M H⁺ ion.

pH = -log (1.8) = - 0.26

(2) pH = 14 - pOH

pOH = - log [OH^-]

To calculate [OH-] first convert mg/ L top g/L

3.06 mg/L = 0.00306 g/L

Strength = Molarity x molecular mass

Molarity = strength / Molecular mass

Molecular mass of Ba(OH)2 = 171.34 g/mol

M = 0.00306 / 171.34 = 1.78 x 10^-5 M

Each molecule of Ba(OH)₂ will give two molecule of OH^-

Ba(OH)2 Ba²⁺ + 2OH^-

Therefore multiply molarity by 2: 2 x 1.78 x 10-5 = 3.56 x 10^-5 M

pOH = -log (3.56 x 10^-5) = -log(3.56) + 5log(10) = 5 - 0.55 = 4.44

pH = 14 - 4.44 = 9.55

(3) First calculate H⁺ ion concentration

pH = -log [H+]

[H+] = antilog(- pH)   = 10^-2.65 = 0.0022 M

Consider the reaction    HA     H⁺    +    A^-

                       Start   0.0112          0            0

                     Change   -x              x             x

          At Equilibrium 0.0112-x    x              x

x = [H+] = [A-] = 0.0022 M

0.0112 - x = [HA] = 0.0112 - 0.0022 = 0.009 M

Ka = [H+] [A-] / [HA] = (0.0022)² / 0.009 = 0.00053 = 5.3 x 10^-4