# 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.

• 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.
– user15489
Commented Sep 7, 2015 at 21:38
• It might help to start think about which glutamate species could exist. Commented Sep 7, 2015 at 21:39
• 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.
– Macy
Commented Sep 7, 2015 at 21:42
– user15489
Commented Sep 7, 2015 at 21:43
• What do you mean by other species? Commented Sep 9, 2015 at 10:36

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