Limiting fraction of association Show that the limiting fraction of association of a base in water, as
  the concentration of base approaches 0, is $$\alpha=\frac{10^7 K_b}{(1+10^7\times K_\mathrm{b})}.$$


I have tried this problem for a long time and have made little progress. Here is what I have tried:

We know that:

\begin{align}
K_\mathrm{b} &amp;= \frac{\ce{[HB+]}\ce{[OH-]}}{\ce{[B]}}\tag{definition}\\
\ce{[HB+]} + \ce{[H+]} &amp;= \ce{[OH-]} \tag{charge balance}\\
\ce{[H+][OH-]} &amp;= 10^{-14} \tag{$K_\mathrm{w}$ at $\pu{25^o C}$}\\
\text{initial concentration of base added} &amp;= \ce{[B]} + \ce{[HB+]} \tag{mass balance}
\end{align}

I know that the limiting fraction of association is 
$$\alpha = \frac{\ce{[HB+]}}{\text{initial concentration of base added}} = \frac{\ce{[OH-]}}{(\ce{[B]} + \ce{[HB+]})},$$ 
but I'm confused about how to proceed from here. I tried putting $\alpha$ in terms of $\ce{[OH-]}$ to get a cancellation, but that proved difficult. How do I proceed?
Start from 

$$K_\mathrm{b} = \frac{\ce{[HB+]}\ce{[OH-]}}{\ce{[B]}}$$

Define 

$$\alpha = \frac{\ce{[HB+]}}{\ce{[B]} + \ce{[HB+]}} $$

or

$$\frac{\ce{[HB+]}}{\ce{[B]}}=\frac{\alpha}{1-\alpha}$$

so that 

$$K_\mathrm{b} = \frac{\alpha\ce{[OH-]}}{1-\alpha}$$

or 

$$\alpha = \frac{K_\mathrm{b}}{K_\mathrm{b}+\ce{[OH-]}}$$

But in the limit of infinite dilution the concentration of $\ce{OH^-}$ is solely determined by the autodissociation of water (if the solution is pure unbuffered water) and

$$\ce{[OH-]} = \sqrt{K_\mathrm{w}}$$

so that 

$$\alpha = \frac{K_\mathrm{b}}{K_\mathrm{b}+K_\mathrm{w}^{1/2}}$$

But at RTP $K_\mathrm{w}^{1/2}\approx 10^{-7}$ resulting in the final expression

$$\alpha = \frac{10^{7}K_\mathrm{b}}{10^{7}K_\mathrm{b}+1}$$