How many grams of sodium dihydrogen phosphate (NaH2PO4, M=120 g/mol) must be added to the solution containing 0.05 mole of disodium hydrogen phosphate (Na2HPO4) in order to produce,

How many grams of sodium dihydrogen phosphate (NaH2PO4, M=120 g/mol) must be added to the solution containing 0.05 mole of disodium hydrogen phosphate (Na2HPO4) in order to produce, after dilution with water to the final volume of 1 L, a phosphate buffer of pH = 7.612? (Non-anonymous question@) (2 Points) Dissociation constants for the phosphoric acid: K1 = 6.3 x 10^-3; K2 = 6.2 x 10^-8; K3 = 5.0 x 10^-13.

The Correct Answer and Explanation is:

To solve this problem, we need to apply the Henderson-Hasselbalch equation:pH=pKa+log⁡([base][acid])\text{pH} = \text{p}K_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)pH=pKa​+log([acid][base]​)

We are given:

• pH = 7.612
• The base is HPO42−\text{HPO}_4^{2-}HPO42−​ (from Na₂HPO₄), which is the conjugate base of H2PO4−\text{H}_2\text{PO}_4^{-}H2​PO4−​
• The acid is H2PO4−\text{H}_2\text{PO}_4^{-}H2​PO4−​ (from NaH₂PO₄)
• Moles of base = 0.05 mol
• Final volume = 1 L, so mol = mol/L

We are dealing with the second dissociation of phosphoric acid:H2PO4−⇌HPO42−+H+\text{H}_2\text{PO}_4^- \rightleftharpoons \text{HPO}_4^{2-} + \text{H}^+H2​PO4−​⇌HPO42−​+H+

For this equilibrium:

• pKa=−log⁡(6.2×10−8)≈7.21\text{p}K_a = -\log(6.2 \times 10^{-8}) \approx 7.21pKa​=−log(6.2×10−8)≈7.21

Step 1: Use Henderson-Hasselbalch Equation

7.612=7.21+log⁡(0.05[acid])7.612 = 7.21 + \log \left( \frac{0.05}{[\text{acid}]} \right)7.612=7.21+log([acid]0.05​)

Subtract 7.21 from both sides:0.402=log⁡(0.05[acid])0.402 = \log \left( \frac{0.05}{[\text{acid}]} \right)0.402=log([acid]0.05​)

Now take the inverse log:100.402≈2.5310^{0.402} \approx 2.53100.402≈2.530.05[acid]=2.53\frac{0.05}{[\text{acid}]} = 2.53[acid]0.05​=2.53

Now solve for the concentration of the acid:[acid]=0.052.53≈0.01976 mol[\text{acid}] = \frac{0.05}{2.53} \approx 0.01976\ \text{mol}[acid]=2.530.05​≈0.01976 mol

Step 2: Convert to grams

We now know we need approximately 0.01976 mol of sodium dihydrogen phosphate (NaH₂PO₄). Use its molar mass to get grams:mass=0.01976 mol×120 g/mol=2.3712 g\text{mass} = 0.01976\ \text{mol} \times 120\ \text{g/mol} = 2.3712\ \text{g}mass=0.01976 mol×120 g/mol=2.3712 g

Final Answer:

2.37 g of NaH2PO4\boxed{2.37\ \text{g of NaH}_2\text{PO}_4}2.37 g of NaH2​PO4​​

Explanation

To prepare a phosphate buffer at pH 7.612, you need a specific ratio of the conjugate base HPO42−\text{HPO}_4^{2-}HPO42−​ and its conjugate acid H2PO4−\text{H}_2\text{PO}_4^-H2​PO4−​. This ratio is calculated using the Henderson-Hasselbalch equation, which relates pH, pKa, and the ratio of base to acid. Since the relevant equilibrium is the second dissociation of phosphoric acid (Ka₂), we use its value Ka=6.2×10−8K_a = 6.2 \times 10^{-8}Ka​=6.2×10−8, and compute pKa=7.21\text{p}K_a = 7.21pKa​=7.21.

The goal is to make the ratio of base to acid equal to the ratio that yields pH 7.612. Plugging into the equation, we find that the base to acid ratio must be about 2.53. With 0.05 mol of base already present, we calculate how much acid is needed to achieve this ratio, which turns out to be approximately 0.01976 mol.

We then convert the required moles of acid to grams using its molar mass (120 g/mol). Multiplying gives us 2.37 grams. This means that, to achieve the target pH in a 1-liter buffer solution, we must add 2.37 grams of sodium dihydrogen phosphate to the solution already containing 0.05 mol of disodium hydrogen phosphate.

This buffer resists changes in pH by neutralizing added acids or bases, making it useful in many chemical and biological processes.