Problem
The phrase predominant charge at ph 7 alanine asks for the net charge of the dominant ionic form of alanine in aqueous solution at pH 7.
Key acid–base data for alanine
Alanine has two relevant ionizable groups (the side chain –CH3 is non-ionizable in water):
| Group | Acid/base pair | Typical \(pK_a\) (assumed) | Charge when protonated | Charge when deprotonated |
|---|---|---|---|---|
| Carboxyl (α-COOH) | \(\mathrm{COOH \rightleftharpoons COO^- + H^+}\) | \(pK_{a1} \approx 2.34\) | \(0\) | \(-1\) |
| Ammonium (α-NH3+) | \(\mathrm{NH_3^+ \rightleftharpoons NH_2 + H^+}\) | \(pK_{a2} \approx 9.69\) | \(+1\) | \(0\) |
Minor variations in published \(pK_a\) values occur; the dominant qualitative conclusion at pH 7 is unchanged.
Step 1: Compare pH to each \(pK_a\)
- For the carboxyl group: pH 7 is far above \(pK_{a1} \approx 2.34\), so the carboxyl is predominantly COO−.
- For the ammonium group: pH 7 is below \(pK_{a2} \approx 9.69\), so the amino group is predominantly NH3+.
Step 2: Quantify with Henderson–Hasselbalch ratios
For a conjugate pair \(\mathrm{HA/A^-}\), the Henderson–Hasselbalch relation is:
\[ \mathrm{pH} = pK_a + \log\!\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right) \]
Carboxyl group (\(\mathrm{COOH/COO^-}\))
\[ \log\!\left(\frac{[\mathrm{COO^-}]}{[\mathrm{COOH}]}\right) = \mathrm{pH} - pK_{a1} = 7.00 - 2.34 = 4.66 \]
\[ \frac{[\mathrm{COO^-}]}{[\mathrm{COOH}]} = 10^{4.66} \approx 4.57 \times 10^{4} \]
The deprotonated fraction is \[ f(\mathrm{COO^-}) = \frac{10^{4.66}}{1 + 10^{4.66}} \approx 0.99998 \]
Ammonium group (\(\mathrm{NH_3^+/NH_2}\))
\[ \log\!\left(\frac{[\mathrm{NH_2}]}{[\mathrm{NH_3^+}]}\right) = \mathrm{pH} - pK_{a2} = 7.00 - 9.69 = -2.69 \]
\[ \frac{[\mathrm{NH_2}]}{[\mathrm{NH_3^+}]} = 10^{-2.69} \approx 2.04 \times 10^{-3} \]
The protonated fraction is \[ f(\mathrm{NH_3^+}) = \frac{1}{1 + 10^{-2.69}} \approx 0.9980 \]
Predominant charge at pH 7 for alanine
The dominant microstate at pH 7 combines the most common states of both groups: NH3+–CH(CH3)–COO−. This is the zwitterion, and its net charge is 0 \((+1) + (-1) = 0\).
A useful check is the isoelectric point for a neutral side-chain amino acid: \[ pI \approx \frac{pK_{a1} + pK_{a2}}{2} = \frac{2.34 + 9.69}{2} = 6.015 \]
Since pH \(= 7.00\) is slightly above \(pI\), the average net charge is slightly negative, but the predominant form remains the neutral zwitterion (net 0).
Visualization: qualitative net charge vs pH for alanine
Common pitfalls
- Confusing “predominant species” (most abundant form) with “average net charge” (a weighted average over all forms).
- Assigning an ionizable side chain to alanine; the –CH3 group does not ionize in water.
- Forgetting that the amino group is typically protonated (NH3+) at neutral pH.