physical chemistry - Calculating charge on amino acid from pKa

I have been given $\mathrm pK_\mathrm a$ values of an amino group, a carboxyl group and a side chain of cysteine. How can I find the ionic charge on it at different $\mathrm{pH}$ values?

Answer

For this, you can use the Henderson-Hasselbalch equation. Using the degree of dissociation, $\alpha$, this can be written as $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\alpha}{1- \alpha}$$ Rewriting to solve for $\alpha$: $$\alpha = \frac{1}{10^{\mathrm{p}K_\mathrm{a} - \mathrm{pH}} + 1}$$ As stated above, $\alpha$ is the degree of dissociation, meaning the degree at which $\ce{H+}$ is dissociated from the ionizable group. At $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$, $\alpha = 0.5$, meaning that $50\%$ of the ionizable groups in question are deprotonated. If $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + 1$, about $90\%$ of all groups are deprotonated and if $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + 2$, about $99\%$ of all groups are deprotonated. (If you want to, take a look at the shape of the $\alpha$ distribution for $\ce{-SH}$ in cysteine at Wolfram Alpha.)

From a textbook I found the following $\mathrm{p}K_\mathrm{a}$ values for cysteine: \begin{align} \mathrm{p}K_\mathrm{a}(\ce{-COOH}) &= 1.9\\ \mathrm{p}K_\mathrm{a}(\ce{-NH_3+}) &= 8.35\\ \mathrm{p}K_\mathrm{a}(\ce{-SH}) &= 10.5 \end{align}

From these values, $\alpha$ can be calculated for each ionizable group at the desired pH and this will give you the net charge of the amino acid. Upon deprotonation, the following changes in charge occur for the ionizable groups: \begin{array}{lclcr} \ce{-COOH} &:& 0 &\rightarrow &-\\ \ce{-NH_3+}&:& + &\rightarrow &0\\ \ce{-SH} &:& 0 &\rightarrow &-\\ \end{array}

As an example, let's calculate the charge of cysteine at pH 10. Using the Henderson-Hasselbalch in a spreadsheet yields the following $\alpha$ values: \begin{array}{lcr} \ce{-COOH} &:& 0.9999\\ \ce{-NH_3+} &:& 0.9781\\ \ce{-SH} &:& 0.2403\\ \end{array}

This gives the following total charge for cysteine at pH 10: $$(-1)\cdot0.9999+(+1)\cdot(1-0.9781)+(-1)\cdot0.2403=-1.218$$