Question

I’m stuck on part B

Armonia, NH, is a weak base with a Kvalue of 1.8 x 10-5 - Part A sher What is the pH of a 0.165 Marmonia solution? Express your answer numerically to two decimal places ► View Available Hints) pH = 11.24 bases, the greater the Previous Ans Correct Significant Figures Feedback: Your answer 11.02 was either rounded differently or used a different number of significant figures than required for this part Part B What is the percent ionization of ammonia at this concentration? Express your answer with the appropriate units. View Available Hints) ionization - 1.732 % Submit Previous Answers X Incorrect; Try Again; 2 attempts remaining Review your calculations, you may have made around the worl

Answer 1

Part A :

Let α be the dissociation of the weak base
                            BOH <---> B ⁺ + OH^-

initial conc.            c               0         0

change               -cα            +cα      +cα

Equb. conc.         c(1-α)        cα      cα

Dissociation constant, K_b = (cα x cα) / ( c(1-α)               

                                          = c α² / (1-α)

In the case of weak bases α is very small so 1-α is taken as 1

So K_b = cα²

==> α = √ ( K_b / c )

Given K_b = 1.8x10^-5

          c = concentration = 0.165 M

Plug the values we get α = √ [(1.8x10^-5)/0.165] = 0.0104
So the concentration of [OH^-] = cα

                                           = 0.165 x0.0104
                                           = 1.72 x 10^-3 M

pOH = - log [OH^-]

        = - log  (1.72 x 10^-3)

        = 2.76

So pH = 14 - pOH

         = 14-2.76

        = 11.24

Part B :

% dissociation = α * 100 = 0.0104* 100 = 1.04 %