Which of the following aqueous solutions has the highest [OH– ]? A. A solution with a pH of 3.0 B. 1 × 10–4 M solution of HNO3 C. A solution with a pOH of 12.0 D. Pure water E. 1 × 10–3 M solution of NH4Cl

The solution that has the highest [OH-] is pure water.

For the solution of pH 3.0;

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 3.0 = 11.0

Since;

pOH = -log[OH-]

[OH-] = Antilog (-pOH)

[OH-] = Antilog (-11.0)

[OH-] = 1 × 10^-11 M

For the 1 × 10–4 M solution of HNO3;

[OH-] [H^+] = 1 × 10–14

[OH-] = 1 × 10–14/[H^+]

[OH-] =  1 × 10–14/ 1 × 10–4

[OH-] = 1 × 10–10 M

For the solution of pOH = 12

[OH-] = Antilog (-pOH)

[OH-] =  Antilog (-12)

[OH-] =  1 × 10–12 M

Pure water has OH- concentration of   1 × 10–7 M

For 1 × 10–3 M solution of NH4Cl;

Since NH4Cl yields the acid HCl in solution;

[H+] = 1 × 10–3 M

[OH-] =  1 × 10–14/ 1 × 10–3 M

[OH-] = 1 × 10–11

The solution that has the highest [OH-]  is  pure water..

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D. Pure water; [OH-] = 1e-7

Explanation:

The greater the concentration, the greater the basicity.

A. A solution with a pH of 3.0

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 3 = 11

pOH = -log[OH-]

[OH-] = Antilog(-pOH)

[OH-] = Antilog(-11)

[OH-] = 1e-11

B. 1 × 10–4 M solution of HNO3

pH = -log[H+]

log10{0.0001} = log10(10−4)=−4, by definition of the log function.

Thus pH = 4,

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 4 = 10

pOH = -log[OH-]

[OH-] = Antilog(-pOH)

[OH-] = Antilog(-10)

[OH-] = 1e-10

C. A solution with a pOH of 12.0

pOH = -log[OH-]

[OH-] = Antilog(-pOH)

[OH-] = Antilog(-12)

[OH-] = 1e-12

D. Pure water

Pure water is considered to neutral and the hydronium ion concentration is 1.0 x 10-7 mol/L which is equal to the hydroxide ion concentration.

[OH-] = 1e-7

E. 1 × 10–3 M solution of NH4Cl

NH4Cl dissolves in solution to form ammonium ions NH+4 which act as a weak acid by protonating water to form ammonia, NH3(aq) and hydronium ions H3O+(aq):

NH+4(aq) + H2O(l) → NH3(aq) + H3O+(aq)

Ka × Kb = 1.0×10−14 assuming standard conditions.

So, Ka(NH+4) = 1.0×10−14 / 1.8×10−5 = 5.56×10−10

Plug in the concentration and the Ka value into the expression:

Ka = [H3O+] × [NH3] / [NH+4]

5.56×10−10 ≈ [H3O+] × [NH3] / [0.001]

5.56×10−13 = [H3O+]^2

(as we can assume that one molecule hydronium must form for every one of ammonia that forms. Also, Ka is small, so x≪0.1.)

[H3O+] = 7.46×10−7

pH=−log[H3O+]

pH = −log(7.45×10−6)

pH ≈ 6.13

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 6.13 = 7.87

pOH = -log[OH-]

[OH-] = Antilog(-pOH)

[OH-] = Antilog(-7.87)

[OH-] = 1.345e-08

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