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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    – 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? $\endgroup$
    Commented Sep 9, 2015 at 10:36

1 Answer 1


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: $$\frac{y}{z} = 10^{2.51}$$


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