You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2HPO4) and you need 1L of the buffer at pH 7.00 with a total phosphate concentration (NaH2PO4 + Na2HPO) of 0.100 M. Hint: Phosphoric acid (H3PO4), a triprotic acid, has 3 pKa values: 2.14,
Equvalents OH 2.0 2.5 3.0 Answer the following questions regarding preparation of this buffer: BIO 101 Spring 2019 (a) Identify acid and its conjugate base in this buffer system. (b) Which pKa value is relevant for the preparation of this buffer? (c) Determine the weight in grams of sodium dihydrogen phosphate (NaH2PO4; 138g/mol) and disodium hydrogen phosphate (Na2HPO4; 142g/mol) needed to prepare this buffer 5. Phosphate buffer (a) What is the pH of a mixture of 0.042 M NaH2P04 and 0.058 M Na2HPO4? (b) If1 mL of 10 M NaOH is added to a liter of the buffer prepared in (a), how much will the pH change? (e) If1 mL of 10 M NaOH is added to a liter of pure water at pH 7, what is the final pH? Compare this with the answer in (b)
The Correct Answer and Explanation is :
(a) Acid and Conjugate Base in the Buffer System
In this buffer system, we are working with the phosphate buffer system. The acid and its conjugate base are:
- Acid: NaH₂PO₄ (Sodium dihydrogen phosphate)
- Conjugate Base: Na₂HPO₄ (Disodium hydrogen phosphate)
In this system, NaH₂PO₄ (H₂PO₄⁻) acts as the acid, and Na₂HPO₄ (HPO₄²⁻) is its conjugate base. The acid dissociates to release H⁺ ions, while the conjugate base accepts these ions to neutralize excess acid.
(b) Relevant pKa Value for Buffer Preparation
For a buffer with a pH of 7.00, the relevant pKa is the one closest to 7. In the case of phosphate, pKa₂ = 7.2. This corresponds to the dissociation of H₂PO₄⁻ (which is the acid) into HPO₄²⁻ (the conjugate base). This pKa is the key factor for the buffer system because it governs the equilibrium between H₂PO₄⁻ and HPO₄²⁻ that will help maintain the pH.
(c) Determining the Weights of Sodium Dihydrogen Phosphate (NaH₂PO₄) and Disodium Hydrogen Phosphate (Na₂HPO₄)
To prepare a 1 L buffer solution at pH 7.00 with a total phosphate concentration of 0.100 M, you need to determine the molar amounts of NaH₂PO₄ and Na₂HPO₄ required to achieve this buffer. The required ratio of NaH₂PO₄ to Na₂HPO₄ is determined using the Henderson-Hasselbalch equation:
[
\text{pH} = \text{pKa} + \log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right)
]
Substitute pH = 7.00, pKa = 7.2, and solve for the ratio of base to acid:
[
7.00 = 7.2 + \log \left(\frac{[\text{Na}_2\text{HPO}_4]}{[\text{NaH}_2\text{PO}_4]}\right)
]
[
\log \left(\frac{[\text{Na}_2\text{HPO}_4]}{[\text{NaH}_2\text{PO}_4]}\right) = -0.2
]
[
\frac{[\text{Na}_2\text{HPO}_4]}{[\text{NaH}_2\text{PO}_4]} = 10^{-0.2} \approx 0.63
]
From the ratio, we know that the molarity of Na₂HPO₄ should be 0.63 times the molarity of NaH₂PO₄. Let:
[
[\text{Na}_2\text{HPO}_4] = x \quad \text{and} \quad [\text{NaH}_2\text{PO}_4] = 0.63x
]
Given that the total concentration is 0.100 M, we can write:
[
x + 0.63x = 0.100
]
[
1.63x = 0.100
]
[
x = \frac{0.100}{1.63} \approx 0.0613 \, \text{M}
]
Thus, the molarity of Na₂HPO₄ (conjugate base) is 0.0613 M, and the molarity of NaH₂PO₄ (acid) is:
[
0.63 \times 0.0613 \approx 0.0386 \, \text{M}
]
Now, calculate the amounts of NaH₂PO₄ and Na₂HPO₄ required for 1 liter (since molarity is moles per liter, the total moles needed are just the molarity times the volume):
- NaH₂PO₄: ( 0.0386 \, \text{mol/L} \times 1 \, \text{L} = 0.0386 \, \text{mol} )
- Na₂HPO₄: ( 0.0613 \, \text{mol/L} \times 1 \, \text{L} = 0.0613 \, \text{mol} )
Finally, using the molar masses of NaH₂PO₄ (138 g/mol) and Na₂HPO₄ (142 g/mol), the required masses are:
- NaH₂PO₄: ( 0.0386 \, \text{mol} \times 138 \, \text{g/mol} = 5.32 \, \text{g} )
- Na₂HPO₄: ( 0.0613 \, \text{mol} \times 142 \, \text{g/mol} = 8.70 \, \text{g} )
So, to prepare 1 L of the buffer at pH 7.00, you need 5.32 g of NaH₂PO₄ and 8.70 g of Na₂HPO₄.
Additional Questions:
- (a) To calculate the pH of a mixture of 0.042 M NaH₂PO₄ and 0.058 M Na₂HPO₄, apply the Henderson-Hasselbalch equation with pKa = 7.2:
[
\text{pH} = 7.2 + \log \left(\frac{0.058}{0.042}\right)
]
[
\text{pH} = 7.2 + \log(1.381) \approx 7.2 + 0.141 \approx 7.34
]
So, the pH is approximately 7.34.
- (b) If 1 mL of 10 M NaOH is added to a liter of the buffer, it will slightly increase the pH. The addition of NaOH will neutralize some of the H₂PO₄⁻ ions (the acid), shifting the buffer equilibrium. However, because this is a buffer solution, the pH change will not be large.
- (c) If 1 mL of 10 M NaOH is added to pure water at pH 7, the pH will increase significantly, as water does not have a buffer to resist the change. You can use the amount of NaOH added to calculate the number of moles of OH⁻ and determine the final pH.
In comparison, adding NaOH to the buffer solution in part (b) results in a smaller pH change, showing the buffering capacity of the system.