what concentration of ammonia is required to give the solution of pH 11.89 (Kb of NH3=1.80*10^-5)?

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what concentration of ammonia is required to give the solution of pH 11.89 (Kb of NH3=1.80*10^-5)?

2007-05-18 06:00:45 · 2 answers · asked by mgp_chem88 1 in Science & Mathematics ➔ Chemistry

2 answers

Kb = [NH4(+)][OH-]/[NH3]

If pH = 11.89, then pOH = 2.11 and [OH-] = 10^(-2.11)
which is equal to 0.0078 mol/L

The concentration of OH- is the same as the concentration of NH4(+)

1.8E-5 = (0.0078)^2 divided by [NH3]

[NH3] = 3.35 M

2007-05-18 06:07:57 · answer #1 · answered by mrfarabaugh 6 · 0⤊ 0⤋

2.318 x 10^-3 M

2007-05-18 06:11:37 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋