Question

Determine the pH of (a) 0.20 M sulfuric acid (b) 0.20 M hydrochloric acid, (c) 0.20 M sodium hydroxide, (d) 0.20 M acetic acid, and (e) 0.20 M solution of NH₃ at 25 ^oC.

Answer 1

Determine the pH of

(a) 0.20 M sulfuric acid

Solution :- H2SO4 is strong acid therefore it gives 2 H^+

So the concentration of the H^+ = 0.20 M * 2 = 0.40 M

Lets calculate the pH

pH = - log [H+]

pH= -log[0.40]

pH= 0.40

(b) 0.20 M hydrochloric acid,

Solution :- HCl is strong acid therefore dissociates completely to give 0.20 M H+

Therefore

pH = - log [H+]

pH= -log[0.20]

pH= 0.70

(c) 0.20 M sodium hydroxide,

Solution :- NaOH is strong base therefore it gives 0.20 M OH-

So pOH = -log [OH-]

pOH =- log [0.20]

pOH = 0.70

pH+ pOH = 14

therefore pH= 14 – pOH

                      = 14 – 0.70

                      = 13.3

(d) 0.20 M acetic acid,

Solution :-

Acetic acid (CH3COOH) is weak acid its ka = 1.8*10^-5

Therefore lets calculate the concentration of the H+ using the ka

Ka = [H+] [CH3COO-]/[CH3COOH]

1.8*10^-5 = [x][x]/0.20 M

1.8*10^-5 * 0.20 = x^2

3.6*10^-6 = x^2

Taking square root of both sides we get

1.89*10^-3 = x = [H+]

Now lets calculate the pH

pH = - log [H+]

pH= -log[0.00189]

pH= 2.72

(e) 0.20 M solution of NH₃ at 25 ^oC.

Solution :-

NH3 is weak base its kb = 1.8*10^-5

Lets calculate the OH- concentration using the kb

Kb= [OH-][NH4+] /[NH3]

1.8*10^-5 = [x][x]/0.20 M

1.8*10^-5 * 0.20 = x^2

3.6*10^-6 = x^2

Taking square root of both sides we get

1.89*10^-3 = x = [OH- ]

Lets calculate the pOH

pOH= -log [OH-]

pOH = -log [1.89*10^-3]

pOH = 2.72

pH+ pOH = 14

therefore

pH= 14 – pOH

    = 14 – 2.7

    = 11.3