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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.

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    $\begingroup$ 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
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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

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