1. a) A solution of 1 x 10 -8 M KOH (aq) is prepared. What is the pH of this sol

Question

1.

a) A solution of 1 x 10-8 M KOH (aq) is prepared. What is the pH of this solution?

b) A phthalate buffer is prepared in which [hydrogen phthalate]= 0.047 M and [phthalic acid] = 0.083 M. If the Ka1 of phthalic acid acid is 1.12 x 10-3, and the Ka2 = 3.91 x 10-6, what is the expected pH of the buffer?

c) The pH of a solution formed from a weak monoprotic acid (HA) is 3.1. If the formal concentration of the acid is 0.1M, what is the acids Ka value?

d) The pH of a saturated solution of a metal hydroxide (with formula M(OH)2) is 9.85. What is the Ksp of M(OH)2?

Explanation / Answer

a)

KOH = [OH-] = 1.0 x 10-8 M

as it is very very dilute we should consider OH- water also

[OH-] from water = 1.0 x 10-7 M

total [OH-] = 1.0 x 10-8 + 1.0 x 10-7

[OH-] = 10-7 [1.1] M

pOH = - log [OH-]

pOH = - log [1.1 x 10-7]

pOH = 6.96

pH = 14 - 6.96

pH = 7.04

b) pH = pKa + log [hydrogen phthalate] / [phthalic acid]

pKa = - log Ka = - log [1.12 x 10-3]

pKa = 2.95

pH = 2.95 + log [0.047] / [0.083]

pH = 2.70

c) pH = 3.1

[H+] = 10-pH = 10-3.1 = 0.00079 M

[A-] = [H+] = 0.00079 M

at equilibrium [HA] = 0.1 - 0.00079 = 0.09921 M

Ka = [H+] [A-] / [HA]

Ka = [0.00079][0.00079] / [0.09921]

Ka = 6.29 x 10-6

d) pH = 9.85

pOH = 14 - 9.85 = 4.15

[OH-] = 10-pOH = 10-4.15

[OH-] = 7.08 x 10-5 M

M(OH)2 <-------------> M+2 + 2OH-

Ksp = [M+2] [OH-]2

[OH-] = 7.08 x 10-5 M

[M+2] = 7.08 x 10-5 / 2 = 3.54 x 10-5 M

Ksp = [3.54 x 10-5] [7.08 x 10-5]2

Ksp = 1.77 x 10-13

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