11. The mass balance equation for F in a 0.02 M aqueous solution of KF is: (B) [
ID: 695127 • Letter: 1
Question
11. The mass balance equation for F in a 0.02 M aqueous solution of KF is: (B) [HFI+ IF1+K1- 0.02 M (C) HFI+IF1+K]+H1+IOH]-0.02 M (D) [F] = 0.02 M (E) [HF] + [F] = 0.02 M 12. Which acid has the weakest conjugate base? (A) HC2Hs02 (B) HF (C) HBr (D) HNO2 (E) HCIO 13. The chemical equation whose equilibrium constant is defined as the K, for the trimethylammonium ion, (CHs)2NH, is: (A) (CH):N + H2O = (CH):NH' + OH. (B) (CHs)N+ H2O (CHs)&NOH;+H (C) (CHs)&NH;+ H2O(CH)N+H,o 14. The pH of a 0.20 M solution of ammonium chloride (pk, for NH4 is 9.245) is: (A) 4.97 (B) 9.03 (C) 0.70 (D) 2.72 (E) 7.00Explanation / Answer
Q11
mass balance must include:
we know
[F] initial = [KF] = 0.02 M
then
[F-] = [HF] + [F-] = 0.02
choose:
E
Q12
weqakest conjguate base --> strongest weak acid
then, form the list
HBr --> strongest acid listed, therefore, its conjguate base will be very weak
Q13
Ka acidicty
so
(CH3)2NH+ = must act as an acid
(CH3)2NH+ = (CH3)2N ; H+
(CH3)2NH+ <--> (CH3)2NH(aq) + H+(Aq)
Q14
Let B --> NH3 and BH+ = NH4+ + for simplicity
the next equilibrium is formed, the conjugate acid and water
BH+(aq) + H2O(l) <-> B(aq) + H3O+(aq)
The equilibrium is best described by Ka, the acid constant
Ka by definition since it is an base:
Ka = [H3O][B]/[BH+]
Ka can be calculated as follows:
Ka = = (10^-9.25)= 5.55*10^-10
get ICE table:
Initially
[H3O+] = 0
[B] = 0
[BH+] = M
the Change
[H3O+] = 0 + x
[B] = 0 + x
[BH+] = -x
in Equilibrium
[H3O+] = 0 + x
[B] = 0 + x
[BH+] = M - x
substitute in Kb expression
Kb = [HA ][OH-]/[A-]
5.55*10^-10= x*x/(0.2-x)
solve for x
x^2 + Ka*x - M*Ka= 0
solve for x with quadratic equation
x = H3O+ = 1.06*10^-5
[H3O+] =1.06*10^-5
pH = -log([H3O+]) = -log(1.06*10^-5) = 4.97
pH = 4.97
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