Hypochlorous acid HClO Ka:3.0 * 10-8
Phenol C6H5OH (or HC6H5O) Ka:1.3 * 10-10
Hydroxylamine HONH2 Kb:1.1 * 10-8
Determine the pH of each of the following solutions (Ka and Kb values are given) all at 25C
1) 9.00×10−2 M hypochlorous acid.
2)7.9×10−3 M phenol.
3)9.0×10−2 M hydroxylamine.
1.
------- HClO(aq) -------------> H^+ (aq) + ClO^- (aq)
I--------0.09 ----------------------0 -----------------0
C------- - x -------------------------+x ---------------+x
E------ 0.09-x ---------------------+x ---------------+x
Ka = [H^+][ClO^-]/[HClO]
3*10^-8 = x*x/(0.09-x)
3*10^-8(0.09-x) = x^2
x = 5.2*10^-5
[H^+] = x = 5.2*10^-5M
PH = -log[H^+]
= -log(5.2*10^-5)
= 4.284 >>>>answer
2.
------- C6H5OH(aq) -------------> H^+ (aq) + C6H5O^-(aq)
I--------0.0079 ----------------------0 ------------------------0
C------- - x -------------------------+x ----------------------+x
E------ 0.0079-x ---------------------+x ---------------+x
Ka = [H^+][C6H5O^-]/[C6H5OH]
1.3*10^-10 = x*x/(0.0079-x)
1.3*10^-10(0.0079-x) = x^2
x = 1*10^-6
[H^+] = x = 1*10^-6M
PH = -log[H^+]
= -log10^-6
= 6
3.
---------- HONH2(aq) + H2O(l) ---------------> HONH3^+ (aq) + OH^- (aq)
I-----------0.09 ------------------------------------------0 -----------------------0
C--------- -x ----------------------------------------- +x ----------------------- +x
E------- 0.09-x ---------------------------------------- +x ----------------------- +x
Kb = [OH^-][HONH3^+]/[HONH2]
1.1*10^-8 = x*x/(0.09-x)
1.1*10^-8(0.09-x) = x^2
x = 3.145*10^-5
[OH^-] = x = 3.145*10^-5M
POH = -log[OH^-]
= -log(3.145*10^-5)
= 4.5024
PH = 14-POH
= 14-4.5024
= 9.4976 >>>>answer
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