molarity = W/G.M.Wt * volume of solution in L = 2/74.5*2 = 0.0134M NaOCl -------

Question

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

                    = 2/74.5*2   = 0.0134M

NaOCl ----------------------> Na^+ (aq) + ClO^- (aq)

                 ClO^- (aq) + H2O ----------------> HClO(aq) + OH^- (aq)

I               0.0134                                               0                0

C             -x                                                      +x               +x

E             0.0134-x                                          +x                  +x

        Kb   = Kw/Ka

                = 1*10^-14/2.9*10^-8   = 3.45*10^-7

    Kb   = [HClO][OH^-]/[ClO^-]

3.45*10^-7   = x*x/0.0134-x

3.45*10^-7*(0.0134-x)   = x^2

     x = 6.78*10^-5

[OH^-]   = x   = 6.78*10^-5M

    POH   = -log[OH^-]

                = -log6.78*10^-5

               = 4.1687

PH    = 14-POH

         = 14-4.1687   = 9.8313 >>>>answer

Explanation / Answer

What is the pH of a solution that is comprised of 2.0 L of water and 2.0 g of NaOCl? Ka [HOCl = 2.9*10^-8).

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