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:

enter image description here

Can anyone shed some light on this?


3 Answers 3


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.


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

Initial (concentrations)


Change (in concentrations)


Equilibrium (concentrations)


We can write the reaction expression as:


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


And solve for $x$:


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

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

Use the quadratic formula:




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:

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


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

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

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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – 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.

  • $\begingroup$ 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. $\endgroup$
    – ringo
    Apr 18, 2015 at 21:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.