Question

Calculate the pH of the following:

a) A 0.50 M solution of aniline (C₆H₅NH₂)

b) A 0.015 M hydrochloric acid solution

c) A 0.0140 M solution of CA(OH)₂

d) A 0.500 M solution of NaNO₃

e) A 0.05 M Na₂O Solution

Answer 1

a)

aniline is weak base as its pKb is very high pKb = 9.4 (wikipedia)

So, pOH of weak base = (pKb-logC)/2 , here C = 0.5 M so, logC = log(0.5) = -0.301

So, pOH = (9.4-(-0.301))/2 = 4.85

so, pH = 14-pOH = 14-4.85 = 9.15

b)

HCl (hydrochloric acid) is a strong acid and thus dissociates completely to give H+ ions

so, concentration of H+ = concentration of HCl = 0.015 M

so, pH = -log(H+) = -log(0.015) = 1.824

c) Ca(OH)2 is a strong base but its solubility in water is very low, so

Ca(OH)2 -> Ca+2 + 2OH-

s 2s

Ksp = [Ca+2][OH-]² = 5.5 x 10^-6 (Taken from csudh)

So, for molar solubility s we have s*(2s)² = 4s³ = 5.5 x 10^-6 so, s = (1.375)^1/3*10^-2 = 1.112*10^-2 M

which is lower higher than the concentration of Ca(OH)2 we've taken (1.4*10^-2 M) so, precipitate will form and the excess Ca+2,OH- will be present in solid form.

so, concentration of OH- = 2s = 2*1.112*10^-2 M = 2.224*10^-2 M

So, pOH = -log(OH-) = -log(2.224*10^-2) = 1.653 so, pH = 14-pOH = 14-1.653 = 12.347

d)

NaNO3 is a salt of strong base NaOH and strong acid HNO3 so, its a neutral salt and thus pH = 7 for its solution.

NaOH + HNO3 -> NaNO3 + H2O