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 __________
Phosphate Buffer pH Calculation
To determine the pH of the phosphate buffer, we need to identify the acidic and basic components and use the Henderson-Hasselbalch equation.
The buffer is prepared using $Na_2HPO_4$ and $NaH_2PO_4$. The relevant acid-base conjugate pair for the given $pK_a$ (6.86) is $H_2PO_4^-$ (the acid) and $HPO_4^{2-}$ (the base).
- $NaH_2PO_4$ provides the acidic species $H_2PO_4^-$.
- $Na_2HPO_4$ provides the basic species $HPO_4^{2-}$.
Calculate Moles of Buffer Components
First, calculate the number of moles for each salt. Since the buffer volume is 1 litre, the moles calculated will directly correspond to the molar concentrations.
- Moles of $Na_2HPO_4$ (source of base $HPO_4^{2-}$): Mass = 208 g Molar mass = 142 g/mol Moles = $\frac{208 \text{ g}}{142 \text{ g/mol}} \approx 1.4648 \text{ mol}$ So, [Base] concentration $\approx 1.4648 \text{ M}$.
- Moles of $NaH_2PO_4$ (source of acid $H_2PO_4^-$): Mass = 71 g Molar mass = 120 g/mol Moles = $\frac{71 \text{ g}}{120 \text{ g/mol}} \approx 0.5917 \text{ mol}$ So, [Acid] concentration $\approx 0.5917 \text{ M}$.
Apply Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the $pK_a$ of the weak acid and the ratio of the concentrations of the conjugate base and the weak acid:
$pH = pK_a + \log \frac{[\text{Base}]}{[\text{Acid}]}$Substitute the calculated moles (which are equal to molar concentrations in 1 L) and the given $pK_a$:
- $pK_a = 6.86$
- [Base] $= [HPO_4^{2-}] \approx 1.4648 \text{ M}$
- [Acid] $= [H_2PO_4^-] \approx 0.5917 \text{ M}$
Calculation:
$pH = 6.86 + \log \frac{1.4648}{0.5917}$ $pH = 6.86 + \log (2.4756)$ $pH = 6.86 + 0.3937$ $pH \approx 7.2537$The calculated pH is approximately 7.25, which falls within the range of 7.2 to 7.3.
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).
- 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 _________
