Two buffer salts have been provided to you to make 100 mL of phosphate buffer with a total phosphate concentration of (0.03 M). They are:已提供两种缓冲盐，用于制作 100 毫升磷酸盐缓冲液，总磷酸盐浓度为 0.03 M。它们是： Acid: Sodium dihydrogen phosphate monohydrate, (NaH2PO4 · H2O) MW: 137.99 g/mol酸：二氢化磷酸钠一水合物，（NaH 2 PO 4 ·H 2 O）最大重量：137.99 克/摩尔 Salt/Conjugated Base: Disodium hydrogen phosphate (Na2HPO4) anhydrous, MW: 141.96 g/mol盐/共轭碱：无水磷酸二钠（Na 2 HPO4），MW：141.96 g/mol What is the amount of base required (in mg) when [A-]/[HA] is 1.01?当[A]/[HA]为 1.01 时，所需的碱量（以毫克计）是多少？

Question

Two buffer salts have been provided to you to make 100 mL of phosphate buffer with a total phosphate concentration of (0.03 M). They are:已提供两种缓冲盐，用于制作 100 毫升磷酸盐缓冲液，总磷酸盐浓度为 0.03 M。它们是： Acid: Sodium dihydrogen phosphate monohydrate, (NaH2PO4 · H2O) MW: 137.99 g/mol酸：二氢化磷酸钠一水合物，（NaH 2 PO 4 ·H 2 O） 最大重量：137.99 克/摩尔 Salt/Conjugated Base: Disodium hydrogen phosphate (Na2HPO4) anhydrous, MW: 141.96 g/mol盐/共轭碱：无水磷酸二钠（Na 2 HPO4），MW：141.96 g/mol What is the amount of base required (in mg) when [A-]/[HA] is 1.01?当[A]/[HA]为 1.01 时，所需的碱量（以毫克计）是多少？

BlackTom AI · Accepted Answer

Total moles of phosphate = Ctotal × V = 0.03 M × 0.100 L = 0.00300 mol.  
Because [A-]/[HA] = 1.01, let moles of HA = nHA and moles of A- = nA = 1.01·nHA. The sum equals total moles:

Step 1 — Partition the total moles using the ratio:
nHA + nA = nHA + 1.01·nHA = (1 + 1.01)·nHA = 2.01·nHA
Therefore nHA = 0.00300 mol / 2.01 = 0.0014925373 mol (intermediate)

Step 2 — Compute moles of base (A-):
nA = 1.01 × nHA = 1.01 × 0.0014925373 = 0.0015074627 mol (intermediate)

Step 3 — Convert moles of base to mass of Na2HPO4 (base salt):
MW (Na2HPO4) = 141.96 g/mol
mass (g) = nA × MW = 0.0015074627 mol × 141.96 g/mol = 0.2139999 g ≈ 0.2140 g

Step 4 — Convert to mg:
0.2140 g × 1000 mg/g = 214 mg

Why this is correct:
- The total phosphate moles are fixed by concentration and volume (0.00300 mol). The given ratio partitions those moles between HA and A- exactly as shown. Each mole of the anhydrous disodium hydrogen phosphate supplies one mole of A-, so direct mole-to-mole conversion to mass using its molecular weight is valid. No additional stoichiometric factors or hydrates apply for the anhydrous base provided.

Common incorrect approaches:
- Using the ratio as fraction of total directly (e.g., using 1.01/0.03 or forgetting to divide by (1+ratio)).  
- Forgetting to convert grams to mg or using the wrong molecular weight (e.g., using the monohydrate MW for the base).

Conclusion: The mass of base required is 214 mg, which matches the provided answer.

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