Relative levels of the two most prominant glutamate species at pH 4.7 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.
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:
$$\frac{y}{z} = 10^{2.51}$$