In order to prepare 100 mL of a 0.50 M sodium phosphate buffer, pH 7.7, how many grams of sodium phosphate monobasic must you add to your solution? For this buffer, both sodium phosphate monobasic and sodium phosphate dibasic are available.
Use MW = 137.99 g/mol for sodium phosphate monobasic for sodium phosphate. Pka 6.82
The Correct Answer and Explanation is :
To prepare 100 mL of a 0.50 M sodium phosphate buffer at pH 7.7, we need to determine the amount of sodium phosphate monobasic (NaH₂PO₄) required. The solution will contain a mixture of sodium phosphate monobasic and sodium phosphate dibasic (Na₂HPO₄), with the proportions adjusted to achieve the desired pH.
Step 1: Determine the required pH and the relevant buffer components
The pH of the buffer is determined by the ratio of the concentrations of the weak acid (NaH₂PO₄) and its conjugate base (Na₂HPO₄). The pH can be calculated using the Henderson-Hasselbalch equation:
[
pH = pKa + \log \left( \frac{[A^-]}{[HA]} \right)
]
Where:
- ( pKa = 6.82 ) for the phosphate buffer.
- ([A^-]) is the concentration of the conjugate base (Na₂HPO₄).
- ([HA]) is the concentration of the weak acid (NaH₂PO₄).
Rearranging the equation to solve for the ratio of base to acid:
[
\log \left( \frac{[A^-]}{[HA]} \right) = pH – pKa
]
[
\log \left( \frac{[A^-]}{[HA]} \right) = 7.7 – 6.82 = 0.88
]
[
\frac{[A^-]}{[HA]} = 10^{0.88} \approx 7.6
]
Step 2: Set up the equation for total buffer concentration
The total concentration of the buffer is given as 0.50 M. This concentration is the sum of the concentrations of the acid and the base:
[
[A^-] + [HA] = 0.50 \, \text{M}
]
Let ( [HA] = x ), so ( [A^-] = 7.6x ). Substituting into the total concentration equation:
[
x + 7.6x = 0.50
]
[
8.6x = 0.50
]
[
x = \frac{0.50}{8.6} \approx 0.0581 \, \text{M}
]
Thus, the concentration of NaH₂PO₄ (weak acid) is approximately 0.0581 M, and the concentration of Na₂HPO₄ (conjugate base) is ( 7.6 \times 0.0581 \approx 0.4419 \, \text{M} ).
Step 3: Calculate the moles of sodium phosphate monobasic (NaH₂PO₄)
We are preparing 100 mL (0.100 L) of the buffer, so the moles of NaH₂PO₄ required are:
[
\text{moles of NaH₂PO₄} = 0.0581 \, \text{M} \times 0.100 \, \text{L} = 0.00581 \, \text{moles}
]
Step 4: Convert moles to grams
Using the molar mass of sodium phosphate monobasic (NaH₂PO₄), which is 137.99 g/mol, the mass of NaH₂PO₄ required is:
[
\text{mass} = 0.00581 \, \text{moles} \times 137.99 \, \text{g/mol} \approx 0.802 \, \text{grams}
]
Conclusion
To prepare 100 mL of a 0.50 M sodium phosphate buffer at pH 7.7, you need to add approximately 0.802 grams of sodium phosphate monobasic (NaH₂PO₄) to the solution.
I will also generate a visual representation of the calculation process for you.
Here is the visual representation of the calculation for preparing your sodium phosphate buffer. It walks through the steps from using the Henderson-Hasselbalch equation to the final result of adding 0.802 grams of sodium phosphate monobasic. Let me know if you need any further clarifications!
