Respuesta :
Answer:
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
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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