Question

Calculate the pH of a 0.10 M solution of barium hydroxide, Ba(OH)2. Express your answer numerically using two decimal places.

Calculate the pH of a 0.10 M solution of NaOHNaOH. Express your answer numerically using two decimal places.

Calculate the pH of a 0.10 M solution of hydrazine, N2H4. Kb for hydrazine is 1.3×10−61.3×10−6. Express your answer numerically using two decimal places.

Calculate the pH of a 0.10 M solution of hypochlorous acid, HOCl. Ka of HOCl is 3.5×10−8. Express your answer numerically using two decimal places.

Calculate the pH of a 0.10 M solution of HCl. Express your answer numerically using two decimal places.

Answer 1

1.   Ba(OH)2 is strong base

Ba(OH)2(aq) -------------> Ba^2+ (aq) + 2OH^- (aq)

0.1M ------------------------------------------------ 2*0.1M

[OH^-] = 2[Ba(OH)2] = 2*0.1 = 0.2M

POH   = -log[OH^-]

            = -log0.2

            =0.7

PH   = 14-POH

        = 14-0.7   = 13.30>>>>answer

NaOH is strong base

NaOH(aq) -------------> Na^+ (aq) + OH^- (aq)

0.1M ----------------------------------------0.1M

[OH^-] =[NaOH] = 0.1 = 0.1M

POH   = -log[OH^-]

            = -log0.1

            =1

PH   = 14-POH

         = 14-1 = 13.00

3.

-------- N2H4 (aq) + H2O(l) -------------> N2H5^+(aq) + OH^- (aq)

I---------- 0.1------------------------------------------ 0 ---------------- 0

C------- -x ------------------------------------------ +x --------------- +x

E-------- 0.1-x -------------------------------------- +x ----------------- +x

               Kb    = [N2H5^+][OH^-]/[N2H4]

               1.3*10^-6   = x*x/(0.1-x)

              1.3*10^-6(0.1-x) = x^2

                x   = 0.00036

[OH^-]   = x   = 0.00036M

POH   = -log[OH^-]

           = -log0.00036

            = 3.44

PH   = 14-POH

        = 14-3.44

         = 10.56 >>>answer

4.

---------- HOCl(aq) + H2O --------------------> H3O^+ (aq) + ClO^- (aq)

I--------- 0.1--------------------------------------------- 0 ------------------ 0

C-------- -x --------------------------------------------- +x --------------- +x

E------- 0.1-x ----------------------------------------- +x ----------------- +x

               Ka    =   [H3O^+][ClO^-]/[HClO]

              3.5*10^-8   = x*x/(0.1-x)

            3.5*10^-8(0.1-x)    = x^2

           x = 6*10^-5

[H3O^+]   = x = 6*10^-5M

PH   = -log[H3O^+]

        = -log(6*10^-5)

        = 4.22

5.

HCl(aq) -------------------> H^+ (aq) + Cl^- (aq)

0.1M ----------------------- 0.1M

[H^+]   = [HCl]   = 0.1M

PH = -log[H^+]

        = -log0.1

         = 1 .00>>>>answer