Question

1a.

Calculate the pH of a solution that contains 0.0282 g of NaOH in 1.50 L of water.

Report your answer to TWO places past the decimal.

1b.

Calculate the pH of a solution that contains 0.0261 g of Ba(OH)₂ in 2.04 L of water.

Report your answer to TWO places past the decimal.

Answer 1

1a.

molarity   = W/G.M.Wt * volume of solution L

              = 0.0282/(40*1.5)

              = 0.00047M

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

0.00047M ------------------------------ 0.00047M

[OH^-]   = [NaOH]   = 0.00047M

POH = -log[OH^-]

        = -log0.00047

        = 3.328

PH   = 14-POH

       = 14-3.328

       = 10.67 >>>>answer

molarity   = W/G.M.Wt * volume of solution L

              = 0.0281/(171*2.04)

              = 8.1*10^-5M

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

8.1*10^-5M ------------------------------ 2*8.1*10^-5M

[OH^-]   =2 [Ba(OH)2]   = 2*8.1*10^-5M   = 0.000162M

POH = -log[OH^-]

        = -log0.000162

        = 3.79

PH   = 14-POH

       = 14-3.79

       = 10.21 >>>>answer