1. a) A solution of 1 x 10 -8 M KOH (aq) is prepared. What is the pH of this sol
ID: 541330 • Letter: 1
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
1*10^-8 M KOH
KOH is strong base PH value range is 10 to 14.
1*10^-8 M KOH is very diluted solution.
The conc of OH^- is equal to conc of OH^- from base and conc of OH^- from water.
[OH^-] = 10^-8 + 10^-7
= 0.1*10^-7 + 10^-7
= (0.1+1) *10^-7
= 1.1*10^-7 M
POH = -log[OH^-]
= -log1.1*10^-7
= 6.9586
PH = 14-POH
= 14-6.9586 = 7.0414
b. PKa1 = -logka1
= -log1.12*10^-3 = 2.9507
PH = Pka1 + log[hydrogen phthalate]/[phthalic acid]
= 2.9507 + log0.047/0.083
= 2.9507 -0.2469 = 2.7038
c. PH = 1/2Pka -1/2logc
3.1 = 1/2Pka-1/2log0.1
3.1 = 1/2PKa -1/2*-1
3.1+ 1 = 1/2PKa
1/2PKa = 4.1
PKa = 8.2
-logKa = 8.2
Ka = 10^-8.2 = 6.3*10^-9
d. PH = 9.85
POH = 14-PH
= 14-9.85 = 4.15
-log[OH^-] = 4.15
[OH^-] = 10^-4.15 = 7.079*10^-5M
M(OH)2 ----------------> M^+ (aq) + 2OH^-
7.079*10^-5 2*7.079*10^-5
Ksp = [M^+]{OH^-]^2
Ksp = 7.079*10^-5 *(14.158*10^-5)^2 = 1.42*10^-12
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