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

Question

Chemistry

Todd Turner

3 February, 02:02

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

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Answers (2)

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Abril Hatfield · Answer 1

A solution with a pOH of 12.0

Explanation:

A. A solution with a pH of 3.0

pH is 3 which is acidic or contain more of H+.

B. 1 * 10-4 M solution of HNO3

log of (1x10-4) = 4 still acidic (more of H+)

C. A solution with a pOH of 12.0

pOH = 12.0 has more of OH-

D. Pure water

has a balanced H + and OH-

E. 1 * 10-3 M solution of NH4Cl

log of (1x10-3) = 3 which is acidic, having more of OH-

Molly Snyder · Answer 2

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