Calculating Lactic Acid pKa
The Henderson-Hasselbalch equation relates pH, pKa, and the concentrations of a weak acid and its conjugate base:
$ \text{pH} = \text{p}K_a + \log_{10} \left( \frac{[\text{conjugate base}]}{[\text{acid}]} \right) $
Given Information
- pH = $4.8$
- [Lactic Acid] (acid) = $0.01$ M
- [Lactate] (conjugate base) = $0.087$ M
Calculation Steps
Rearrange the Henderson-Hasselbalch equation to solve for pKa:
$ \text{p}K_a = \text{pH} - \log_{10} \left( \frac{[\text{Lactate}]}{[\text{Lactic Acid}]} \right) $
Substitute the given values into the equation:
$ \text{p}K_a = 4.8 - \log_{10} \left( \frac{0.087}{0.01} \right) $
Calculate the ratio of concentrations:
$ \frac{0.087}{0.01} = 8.7 $
Calculate the base-10 logarithm of the ratio:
$ \log_{10}(8.7) \approx 0.9395 $
Calculate the pKa:
$ \text{p}K_a = 4.8 - 0.9395 \approx 3.8605 $
Final Answer
Rounding the calculated pKa to one decimal place:
$ \text{p}K_a \approx 3.9 $
Important Questions from Chemical Biology {p}K_a Henderson Hasselbalch
- The equilibrium dissociation constant of acetic acid is $1.74 \times 10^{-5}$ $M$. The $pK_a$ of acetic acid (rounded off to one decimal place) is ________.
- The ionic product of water at $40$ °C is $2.92 \times 10^{-14}$ $M^2$. The pH of water at $40$ °C is ________ (rounded off to 2 decimal places).
An aqueous solution of aspirin (HA) is prepared at pH 7.4. The ratio of concentration of $A^-$ and HA at equilibrium is ________ (round off to the nearest integer).
Given: $K_a$ of aspirin is $3.98 \times 10^{-4}$
- If 50 mL of 0.02 M HCl is added to 950 mL of $H_2O$, then the pH of the final solution will be _________
One litre of phosphate buffer was prepared by adding 208 grams of $Na_2HPO_4$ (Mol. wt. 142) and 71 grams of $NaH_2PO_4$ (Mol. wt. 120) in water. If the $pK_a$ for the dissociation of $H_2PO_4^-$ into $HPO_4^{2–}$ and $H^+$ is 6.86, the pH of the buffer will be __________
