5
$\begingroup$

In the following examples; I don't understand why $\ce{Na_3PO_4}$ is a much more complicated process when calculating $\mu$ compared to $\ce{NaH_2PO_4}$ and $\ce{Na_2HPO_4}$. Specifically, I don't understand how to determine which species to include in the $\mu$ equation. All of the examples I've come across just behave just using water and the original species at most. What is more confusing also, is why doesn't the $\ce{NaH_2PO_4}$ and $\ce{Na_2HPO_4}$ include the concentrations and charges of their conjugates as the last example with $\ce{Na_3PO_4}$ does.Thanks for any help!!! Here is the work:

$0.1000M$ of each reagent to start.

$\ce{NaH_2PO_4}$

$$\text{Ignoring} ~H^+ \text{and}~ OH^-. [C] << 0.1M$$ $$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [H_2PO_4^-](-1)^2)$$

$$\mu = \frac{1}{2}\sum([0.1M](+1)^2 + [0.1M](-1)^2)$$

$$\mu = 0.1000M$$

$\ce{NaH_2PO_4}$

$$\text{Ignoring} ~H^+ \text{and}~ OH^-. [C] << 0.1M$$ $$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [H_2PO_4^-](-2)^2)$$

$$\mu = \frac{1}{2}\sum([0.2M](+1)^2 + [0.1M](-1)^2)$$

$$\mu = 0.3000M$$

$\ce{Na_3PO_4}$

Without using Quadratic (which is just to keep things simple): $$[OH^-]=0.044M$$ $$[PO_4^{3-}]=0.056M$$ $$[HPO_4^{2-}]=0.044M$$ $$[H^+]=\frac{K_w}{0.044M}=2.27*10^{-13}M$$ $$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [PO_4^{3-}](-3)^2+[HPO_4^{2-}](-2)^2+[H^+](+1^2))$$ $$\mu = \frac{1}{2}\sum([0.3M](+1)^2 + [0.056M](-3)^2+[0.044M](-2)^2+[2.27*10^{-13}M](+1^2))$$ $$\mu=0.5120M$$

$\endgroup$
3
$\begingroup$

The third pKa of phosphoric acid (going from disodium to trisodium phosphate) is 12.32.

When trisodium phosphate is added to water it is a relatively strong base, beginning to titrate H20 to OH-, to a significant degree. In other words, to be above the pKa of 12.32, the concentration of OH- needs to be at least 0.02M.

In the other examples, since the other two pKas of phosphoric acid are 2.15 and 7.2, the pH in those examples will be about midway between 2.15 and 7.2 for the monosodium salt and about midway between 7.2 and 12.32 for the disodium salt, so the concentrations of H+ and OH- are insignificant compare to the other ions.

$\endgroup$
  • $\begingroup$ So if the concentrations of H+/OH- are insignificant. You're saying it's safe to assume that the conjugate is just as insignificant? $\endgroup$ – PiZzL3 Feb 28 '15 at 23:42
  • $\begingroup$ Also, if I were to calculate Mu for H3PO4; then I would include the conjugate since the pKa is ~2.15 or close to 0? $\endgroup$ – PiZzL3 Feb 28 '15 at 23:44
  • $\begingroup$ @PiZzL3 "You're saying it's safe to assume that the conjugate is just as insignificant?" Yes, because one OH- ion is created for each conjugate acid created, when you state with purely the base. "Also, if I were to calculate Mu for H3PO4; then I would include the conjugate since the pKa is ~2.15 or close to 0?" Yes. $\endgroup$ – DavePhD Mar 1 '15 at 0:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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