1. If the pH of a solution of NH 3 is 11.67, what is the concentration of NH 3 i

Question

1.       If the pH of a solution of NH3 is 11.67, what is the concentration of NH3 in the solution? (For NH3, Kb = 1.8x10–5)

(a)   0.00029 M

(b)   0.0038 M

(c)   0.0047 M

(d)   0.82 M

(e)   1.22 M

(f)   2.33 M

2.       What would be the concentration of HPO42– in a 5.0 M H3PO4 solution? pKa1 = 2.16, pKa2 = 7.19, and pKa3 = 12.32 for H3PO4.

(a)   4.8x10–13 M

(b)   6.5x10–8 M

(c) 6.9x10–3 M

(d)   0.19 M

(e)   5.0 M

This is for a practice test, so I know the answers should be 1.22 M, and 6.5*10^-8 by looking at an answer key but I don't know how to get to that answer. Can you show me the steps please? Thank you!

Explanation / Answer

Ammonia is a weak base with Kb = 1.8 × 10–5.

pH = 11.67 that is pOH = 14.00 – pH = 2.33

[OH-] = 10–pOH = 10–2.33 = 4.677 × 10-3 M = 0.0047 M

[OH-] = Kb * { Cweak base - (OH-)} / (OH-)

0.0047 = 1.8 x 10-5 * { Cweak base - 0.0047} / 0.0047

Cweak base = 0.07176 M

So, we have 0.07176 mol ammonia in 1 L solution, that is 1.22 g ammonia (m = n × M.W. = 0.07176 mol × 17 g/mol) in 1 L solution. The concentration is 1.22 g/L.

So answer is (e ) 1.22

Answer(2)

Ka1 = [H+]*[H2PO4-]/[H3PO4]
Ka2 = [H+]*[HPO4^2-]/[H2PO4-]
Ka3 = [H+]*[PO4^-3-]/[HPO4^2-]

Ka1 = 6.918*10^-3
Ka2 = 6.166*10^-8
Ka3 = 4.786*10^-13

Because Ka1 >> Ka2, assume that all H+ comes from the first dissociation:
H3PO4 <---------> H+ + H2PO4-
and since Ka2 >> Ka3,
[H2PO4-] = [H+]
then from the equilibrium of the second dissociation:
Ka2 = [H+]*[HPO4^2-]/[H2PO4-] = [HPO4^2-]
[HPO4^2-] = Ka2 = 6.5*10^-8 M

So answer is (b) 6.5 x 10^-8 M

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