Iterative method for calculating pH of a weak acid / base
This method of repeating iterations is interesting, though I would suggest something perhaps a little more straightforward (and rigorous) than doing something until you are "acceptably close."

This method I learned is called a RICE diagram.

Reaction

$$\ce{PO4_{(aq)}^{3-} +H2O_{(l)}&lt;=&gt;HPO4_{(aq)}^{2-} +OH^{-}_{(aq)}}$$

Initial (concentrations)

$$[\ce{PO4^{3-}}]=0.20~~~~~[\ce{HPO4^{2-}}]=0~~~~~[\ce{OH-}]=0$$

Change (in concentrations)

$$[\ce{PO4^{3-}}]=-x~~~~~[\ce{HPO4^{2-}}]=+x~~~~~[\ce{OH-}]=+x$$

Equilibrium (concentrations)

$$[\ce{PO4^{3-}}]=0.20-x~~~~~[\ce{HPO4^{2-}}]=x~~~~~[\ce{OH-}]=x$$

We can write the reaction expression as:

$$K_\mathrm{b}=\frac{[\ce{HPO4^{2-}}]\cdot[\ce{OH-}]}{[\ce{PO4^{3-}}]}$$

We substitute in the equilibrium concentrations and the $K_\mathrm{b}$:

$$2.22\times10^{-2}=\frac{[x]\cdot[x]}{[0.20-x]}$$

And solve for $x$:

$$2.22\times10^{-2}=\frac{x^2}{0.20-x}$$

$$4.44\times10^{-3}-2.22\times10^{-2}\cdot x=x^2$$

$$0=x^2+2.22\times10^{-2}\cdot x-4.44\times10^{-3}$$

Use the quadratic formula:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

$$x=\frac{-(2.22\times10^{-2})\pm\sqrt{(2.22\times10^{-2})^2-4(1)(-4.44\times10^{-3})}}{2(1)}$$

$$x\approx-7.87\times10^{-2},~5.65\times10^{-2}$$

Obviously the negative solution is not applicable to our problem, so $x=5.65\times10^{-2}$.

Since $x=[\ce{OH-}]$, $[\ce{OH-}]=5.65\times10^{-2}$.

We know that:

$$[\ce{H+}]\cdot [\ce{OH-}]=1\times10^{-14}$$

so:

$$[\ce{H+}] \cdot [5.65\times10^{-2}]=1\times10^{-14}$$

$$[\ce{H+}] =\frac{1\times10^{-14}}{5.65\times10^{-2}}\approx1.77\times10^{-13}$$

If you have access to a calculator with a numerical solver function, I would recommend that instead of using the quadratic formula, but if not, this is how it is done.