# Iterative method for calculating pH of a weak acid / base

The problem asks to determine the $$[\ce{H+}]$$ in a $$0.20~\mathrm{M}$$ solution of $$\ce{Na3PO4}$$. The $$K_\mathrm{a}$$ of $$\ce{HPO4-}$$ was given as $$4.5\times 10^{-13}$$, which then allows one to calculate the corresponding $$K_\mathrm{b}$$ as $$2.22\times10^{-2}$$.

As any acid-base problem, I simply set up the expression $$\frac{x^2}{0.20~\mathrm{M}}=2.22 \times 10^{-2}$$ with $$x$$ being $$[\ce{OH-}]$$ and solved for $$[\ce{H+}]$$.

However, the solution involves an interesting iteration which I have never seen before for this type of problem:

Can anyone shed some light on this?

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

$$\ce{PO4_{(aq)}^{3-} +H2O_{(l)}<=>HPO4_{(aq)}^{2-} +OH^{-}_{(aq)}}$$

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 $x$:

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

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

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

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

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

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

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.

• The RICE table (or ICE table) is pretty much what the book used. However, both the book and the questioner had done the approximation that x is insignificant, which would lead to large deviations. Thus, successive iterations could lead to a close answer. Alternatively, the x could be included as significant and the quadratic could be solved, as is demonstrated here. I think the book was trying to make a point about checking your approximations, but not approximating is just as well, or perhaps even better.
– Andy
Apr 18, 2015 at 21:22

Just to add to what everyone is saying, the reason that the iterations are shown is to arrive at a more accurate answer. Your method was also correct as long as you assume that $$0.20 - x \approx 0.20$$, but the problem's provided solution was just more accurate.

A good way to remember this is that if after you find the concentration of $$\ce{x}$$ (not doing the quadratic method), divide it by the original concentration of $$\pu{0.20 M}$$ and calculate the percent dissociation.

If $$x \leq 5\%$$ then your approximation is acceptable, and you don't have to worry about solving for $$x$$ using the quadratic equation and can cancel out the insignificant $$x$$ in the denominator.

However, if you want a more precise answer, it is better to use the quadratic equation to solve for $$x$$.

The correct equation for determining the value of $$x$$, as noted by ringo, is $$2.22\times10^{-2}=\frac{[x]\cdot[x]}{[0.20-x]}$$ With a $$K$$ value sufficiently low, you can ignore the $$x$$ value on the bottom of the fraction, as it is negligible compared to the initial concentration of the species present ($$0.20$$).

However, in this case the $$K$$ value of $$2.22\times10^{-2}$$ is actually quite close to the initial concentration of the $$\ce{PO4_{(aq)}^{3-}}$$ of $$0.20$$ (only around $$10$$ times less). As a result, we cannot ignore the $$x$$ in the denominator, as that would cause a significant change in the ultimate value of $$x$$. This is shown by the iterations you provided, the initial (approximate) $$x$$ value being about $$\pu{0.01 M}$$ different from the final. The iterations then are just a method of compensating for ignoring the bottom $$x$$ value from the very beginning. I was taught the same method that ringo outlined, and it is the more accurate method, but the iterations seem to work as well.

• I was also taught this, but I didn't like the wishy-washiness of being "sufficiently small." I was more for the method that would give me the right answer no matter the size of the $K_\mathrm{a}$ (or $K_\mathrm{b}$ as the case may be). This method does work for some cases, however, so really I say it comes down to preference. Apr 18, 2015 at 21:20