Question

4.) Which of the following would be the least acidic (5 pts)? 4.0 M HC,H3O2(aq) K, = 1.8 x 10° 3.0 M HCI (aq) mel c. 3.0 M H2SO4(aq) d. 5.5 M HBr

Answer 1

a) Consider the ionization of acetic acid, HC₂H₃O₂ as below.

HC₂H₃O₂ (aq) <=====> H⁺ (aq) + C₂H₃O₂^- (aq)      K_a = 1.8*10^-5

Write down the expression for K_a.

K_a = [H⁺][C₂H₃O₂^-]/[HC₂H₃O₂]

=====> 1.8*10^-5 = (x)(x)/(4.0 – x)

Since K_a is small, hence, we can assume x << 4.0 M and write

1.8*10^-5 = x²/(4.0)

=====> x² = 4.0*1.8*10^-5

=====> x² = 7.2*10^-5

=====> x = 8.48*10^-3

Therefore, [H⁺] = 8.48*10^-3 M

and pH = -log [H⁺] = -log (8.48*10^-3 M)

=====> pH = 2.07 (ans)

b) Consider the ionization of HCl.

HCl (aq) -------> H⁺ (aq) + Cl^- (aq)

As per the stoichiometric equation,

1 M HCl = 1 M H⁺.

pH = -log [H⁺] = -log [HCl]

= -log (3.0 M)

= 3.0 (ans)

c) Consider the ionization of H₂SO₄.

H₂SO₄ (aq) ---------> 2 H⁺ (aq) + SO₄^2- (aq)

As per the stoichiometric equation,

1 M H₂SO₄ = 2 M H⁺

Therefore,

3.0 M H₂SO₄ = 6.0 M H⁺

pH = -log [H⁺] = -log (6.0 M)

= 6.0 (ans).

d) Consider the ionization of HBr.

HBr (aq) -------> H⁺ (aq) + Br^- (aq)

As per the stoichiometric equation,

1 M HCl = 1 M H⁺.

pH = -log [H⁺] = -log [HBr]

= -log (5.5 M)

= 5.5 (ans)

The higher is the pH of the solution, the less acidic is the solution. Since solution (c) 3.0 M H₂SO₄ has the highest pH, hence, (c) 3.0 M H₂SO₄ (aq) is the least acidic (ans).