Calculating Acetic Acid's pKa
The relationship between the acid dissociation constant ($K_a$) and its $pK_a$ is given by the formula:
$ pK_a = -\log_{10}(K_a) $
Acetic Acid pKa Calculation
Given the dissociation constant for acetic acid, $K_a = 1.74 \times 10^{-5}$ $M$. We can calculate the $pK_a$ using the formula:
- Step 1: Substitute the $K_a$ value into the formula:
$ pK_a = -\log_{10}(1.74 \times 10^{-5}) $
- Step 2: Apply the logarithm properties ($\log(a \times b) = \log(a) + \log(b)$ and $\log(10^x) = x$):
$ pK_a = -(\log_{10}(1.74) + \log_{10}(10^{-5})) $
$ pK_a = -(\log_{10}(1.74) - 5) $
- Step 3: Calculate the value of $\log_{10}(1.74)$:
$ \log_{10}(1.74) \approx 0.2405 $
- Step 4: Complete the calculation for $pK_a$:
$ pK_a = -(0.2405 - 5) $
$ pK_a = -(-4.7595) $
$ pK_a \approx 4.7595 $
- Step 5: Round the result to one decimal place as requested:
$ pK_a \approx 4.8 $
The calculated $pK_a$ value for acetic acid, rounded to one decimal place, is 4.8.
Important Questions from Chemical Biology {p}K_a Henderson Hasselbalch
- 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).
- In a lactic acid solution at pH $4.8$, the concentrations of lactic acid and lactate are $0.01$ M and $0.087$ M, respectively. The calculated pKa of lactic acid is ________ (Round off to one decimal place)
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 __________
