3
$\begingroup$

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.

$\endgroup$
5
  • 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
    Commented Sep 7, 2015 at 21:38
  • $\begingroup$ It might help to start think about which glutamate species could exist. $\endgroup$ Commented Sep 7, 2015 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
    Commented Sep 7, 2015 at 21:42
  • $\begingroup$ Please edit the question with your thinking. $\endgroup$
    – user15489
    Commented Sep 7, 2015 at 21:43
  • $\begingroup$ What do you mean by other species? $\endgroup$
    – WYSIWYG
    Commented Sep 9, 2015 at 10:36

1 Answer 1

2
$\begingroup$

The two species are related. They are conjugated acid and base to each other. Assuming one species $\ce{A^2-}$ with concentration $x \,\pu{mol/L}$, the second species $\ce{HA-}$ with concentration $y\,\pu{mol/L}$. There is a third species $\ce{H2A}$ with concentration $z \pu{mol/L}$.

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

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

Based on above two equations, the two most prominant species are $\ce{HA-}$ and $\ce{H2A}$ and ratio should be $y/z$. I utilized definition of $K_\mathrm a$ instead of $\mathrm pK_\mathrm a$ to get the above two equations. $10^{-4.7}$ is simply $[\ce{H+}]$ at $\mathrm{pH} = 4.7$

Therefore the final answer is: $$\frac{y}{z} = 10^{2.51}$$

$\endgroup$

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.