1. [ 3 points] Calculate the pH of the following solutions Use the assumption technique first , then solve using the quadratic if necessary. a 0.1 M solution of a weak acid HA with Ka

1. pH = 3

2. pH = 2.2

Explanation:

Ka = 1 x 10 -5

2. 0.01 M solution of a weak acid HX with Ka = 0.01

1, Ka of HA is written:

Ka = 1x10⁻⁵ = [H+] [A-] / [HA]

HA ⇄ H+ + A-

When 0.1M HA reacts, some will produce H+ and A-, its concentrations will be:

[H+] = X

[A-] = X

[HA] = 0.1M - X

Replacing:

1x10⁻⁵ = [X] [X] / [0.1M - X]

1x10⁻⁶ - 1x10⁻⁵X = X²

1x10⁻⁶ - 1x10⁻⁵X - X² = 0

Solving the quadratic formula:

X = -0.001005M. False solution. There is no negative concentrations

X = 0.000995M. Right solution.

[H+] = 0.000995M

As pH = -log [H+]

2. For HX equilibrium:

HX ⇄ H+ + X-

Ka = 0.01M = [H+] [X-] / [HX]

0.01 = [X] [X] / [0.01M- X]

0.0001 - 0.01X = X²

0.0001 - 0.01X - X²= 0

Solving for X:

X = -0.016M. False solution. There is no negative concentrations.

X = 0.00618M. Right solution.

pH = -log 0.00618

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