Calculate the relative levels (i.e. ratios of concentrations) of the two most prominant glutamate species at pH 4.7.

I know that the Henderson–Hasselbalch equation is to be used which is $$\mathrm{pH=p}K_\mathrm a + \log_{10}\mathrm{\frac{[A^-]}{[HA]}}$$

For one glutamate species, this would change to $$\mathrm{4.7=2.19 + \log_{10}\frac{[A^-]}{[HA]}}$$

and for the other species it would be $$\mathrm{4.7=9.67 + \log_{10}\frac{[A^-]}{[HA]}}$$

I know there are some other variables I need to figure out to solve the equation, but I'm not sure how to go about finding them with the information given.

  • 1
    $\begingroup$ Welcome to Chemistry.SE. Take the tour to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that ‎we aren't doing your homework for you. $\endgroup$ – user15489 Sep 7 '15 at 21:38
  • $\begingroup$ It might help to start think about which glutamate species could exist. $\endgroup$ – pH13 - Yet another Philipp Sep 7 '15 at 21:39
  • $\begingroup$ I know that the Henderson–Hasselbalch equation is to be used which is pH= pKa + log [A-]/[HA]. What I'm first stumped on is how I would go about finding the pKa and then calculating the conjugate base and acid. $\endgroup$ – Macy Sep 7 '15 at 21:42
  • $\begingroup$ Please edit the question with your thinking. $\endgroup$ – user15489 Sep 7 '15 at 21:43
  • $\begingroup$ What do you mean by other species? $\endgroup$ – WYSIWYG Sep 9 '15 at 10:36

The two species are related. They are conjugated acid and base to each other. Assuming one species $A^{2-}$ with concentration $xmol/L$, the second species $HA^{-}$ with concentration $ymol/L$. There is a third species $H_2A$ with concentration $zmol/L$.

$\frac{10^{-4.7}\times x}{y}=10^{-9.67} \ \ (1)$

$\frac{10^{-4.7}\times y}{z}=10^{-2.19} \ \ (2)$

Based on above two equations, the two most prominant species are $HA^-$ and $H_2A$ and ratio should be $y/z$. I utilized definition of $K_a$ instead of $pK_a$ to get the above two equations. $10^{-4.7}$ is simply $[H^+]$ at pH 4.7

| improve this answer | |

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.