acid base - calculating pH - is my answer correct?

I'am working on this exercice, Is my answer correct? and can you please help me to solve the second part?

exercice:

1/Calculate the ionization coefficient and pH (with two methods) of a 0.1M ammonia solution of pkb = 4.7

2/ to one liter of the previous solution, an equal volume of water is added. Calculate the new ionization coefficient and deduce the pH of this new solution

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1/ For the first part my answer is :

$\ce{NH3 + H2O = NH4+ + OH^-}\tag{1}$

$Ka = \dfrac{\ce{[NH4+][OH^-]}}{\ce{[NH3+]}}\tag{2}$

The coefficient of ionization $\alpha$

$\alpha = \sqrt{Ka/C°}\tag{3}$

$Pkb = \log{kb}\tag{4}$

$Kb= 10-pkb\tag{5}$

$\alpha =0,014\tag{6}$

Two methods:

First one:

$pH= \frac{1}{2} (pka + 14 + \log{C°})\tag{7} $

$pka + pkb=14\tag{8}$

$pH = 11,15\tag{9}$

second method:

$14 + \log (\alpha C°) = 11,15\tag{10}$

for the second part, i am lost

Thank you