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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?

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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 $x$:

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

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

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

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.

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  • $\begingroup$ 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 '15 at 21:22
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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.

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  • $\begingroup$ 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 '15 at 21:20
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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 $x$.

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