Question

1.Calculate the pH of a 0.075 M soultion of hydrazoic acid, which is a weak acid. Hydrazoic acid, HN3, has an acid dissociation constant Ka, equal to 1.9 x 10-5. Also, calculate the concentrations of the azide ion (N3-) and the hydroxide ion. (answers are ph=2.93; [N3-]=0.0012M ; [OH-]= 8.4x10-12) I just need help how to get them!

2. Calculate the pH that would result if 15.0 mL of 0.075 M NaOH, strong base, were added to 15.0mL of the solution in the problem above. (my professor got pH=8.65)

Answer 1

1)

H+ ion conc = (Ka*c)^1/2   = (1.9*10^-5*0.075)^1/2   = 1.1937*10^-3

pH = - log(H+ ion conc) = - log(1.1937*10^-3)

                                     = +2.93

Again HN3 →   H+   + N3 -

             1            1         1     mole ratio

            so N3 -   = H+ ion conc   = 1.193*10^-3    = .0012 kg

Now pH + pOH   = 14

pOH   = 14 - 2.93   =   11.07

So - log(OH - ion conc)   =   11.07    taking antilogs  

OH - ion conc   = 8.51 * 10^-12 M

2)

   H+ ion conc = 1.1937 *10^-3 * 15/1000   = 1.79*10^-5      ( because   H+ ion conc   = MV= molarity * vol

                                                                                                                                             in litres

                                                                                 Here litres in 15 ml = 15/1000)

OH - ion conc = 0.075*15/1000   + 8.51* 10^-12                  = 1.125*10^-3

                        ( in 15 ml NaOH sol) ( already present in sol)

So pOH   = - log (OH - ion conc) = - log (1.125* 10^ -3) = 2.948

Now pH + pOH = 14

so pH = 14 - pOH   = 14 - 2.948    = 11.052