Question

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.

Answer 1

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