Question

Calculate the [H ] and pH of a 0.0045 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20 × 10-5. Use the method of successive approximations in your calculations or the quadratic formula.?
[H+] = M?
pH = ?

Answer 1

            N3H(aq) ---------------------> N3^- (aq) + H^+ (aq)

I          0.0045                                0                    0

C           -x                                     +x                   +x

E         0.0045-x                               +x                +x

          Ka    =   [N3^-][H^+]/[N3H]

      2.2*10^-5   = x*x/(0.0045-x)

    2.2*10^-5*(0.0045-x)     =   x^2

   0.000022*0.0045- 0.000022x = x^2

     x^2 +0.000022x-0.00000009 = 0

x   = -bb^2-4ac/2a

x   = -0.000022(0.00002)^2-4*1*0.000022/2*1

x = 0.0003

[H^+]   =x = 0.0003M

PH   = -log[H^+]

          = -log0.0003

PH   = 3.523