The pOH of an aqueous solution of 0.553 M hypochlorous acid is The pH of an aque
ID: 1039125 • Letter: T
Question
The pOH of an aqueous solution of 0.553 M hypochlorous acid is The pH of an aqueous solution of 0.265 M potassium fluoride, KF (?. ?Sw Is this solution acidic, basic, or neutral? The pH of an aqueous solution of 0.246 M sodium cyanide, NaCN (ag), is Is this solution acidic, basic, or neutral? Calculate the hydronium ion concentration in an aqueous solution of 1.09x102 M carbonic acid, H2CO3 (ag). Calculate the concentration of HCOs in an aqueous solution of 0.2430 M carbonic acid, H2COs (aa) [HCOs] The hydroxide ion concentration, [OH], of an aqueous solution of 0.553 M hydroxylamine (a weak M. base with the formula NH2OH),Kb-9.1x109, is: OH] M.Explanation / Answer
HOCl ----------------> H^+ (aq) + OCl^- (aq)
I 0.553 0 0
C -x +x +x
E 0.553-x +x +x
Ka = [H^+][OCl^-]/[HOCl]
3*10^-8 = x*x/0.553-x
3*10^-8 *(0.553-x) =x^2
x = 0.000128
[H^+] = x = 0.000128M
[OH^-] = Kw/{H^+]
= 1*10^-14/0.000128 = 7.8*10^-11M
POH = -log[OH^-]
= -log7.8*10^-11
= 10.1079
KF (aq) -----------------> K^+ (aq) + F^- (aq)
0.265M 0.265M
F^- (aq) + H2O ----------------> HF + OH^-
I 0.265 0 0
C -x +x +x
E 0.265-x +x +x
Kb = Kw/Ka
= 1*10^-14/6.3*10^-4 = 1.58*10^-11
Kb = [HF][OH^-]/[F^-]
1.58*10^-11 = x*x/0.265-x
1.58*10^-11 *(0.265-x) = x^2
x = 2.05*10^-6
[OH^-] = x= 2.05*10^-6M
POH = -log[OH^-]
= -log2.05*10^-6
= 5.6882
PH =14-POH
= 14-5.6882 = 8.3118
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