What mass (in g) of sodium dihydrogen phosphate, NaH2PO4, must be added to 100.0 mL of 0.100 M disodium hydrogen phosphate, Na2HPO4, in order to prepare a buffer with pH = 7.00? Assume the volume of the solution does not change with the addition of solid NaH2PO4.
The correct answer and explanation is:
Correct Answer: 0.719 grams of NaH₂PO₄ must be added.
Explanation:
To solve this problem, we use the Henderson-Hasselbalch equation: pH=pKa+log([base][acid])\text{pH} = \text{p}K_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)
For the phosphate buffer system involving H₂PO₄⁻ (acid) and HPO₄²⁻ (base), the relevant dissociation equilibrium is: H2PO4−⇌HPO42−+H+\text{H}_2\text{PO}_4^- \rightleftharpoons \text{HPO}_4^{2-} + \text{H}^+
The pKa for this equilibrium is 7.20.
Given:
- Target pH = 7.00
- [Base] = [Na₂HPO₄] = 0.100 M
- Volume = 100.0 mL = 0.1000 L
Step 1: Use the Henderson-Hasselbalch equation to find the acid/base ratio. 7.00=7.20+log(0.100[acid])7.00 = 7.20 + \log \left( \frac{0.100}{[\text{acid}]} \right) 7.00−7.20=log(0.100[acid])7.00 – 7.20 = \log \left( \frac{0.100}{[\text{acid}]} \right) −0.20=log(0.100[acid])-0.20 = \log \left( \frac{0.100}{[\text{acid}]} \right) 10−0.20=0.100[acid]10^{-0.20} = \frac{0.100}{[\text{acid}]} 0.631=0.100[acid]0.631 = \frac{0.100}{[\text{acid}]} [acid]=0.1000.631=0.1585 M[\text{acid}] = \frac{0.100}{0.631} = 0.1585\ \text{M}
Step 2: Calculate moles of NaH₂PO₄ needed. moles=M×V=0.1585 mol/L×0.1000 L=0.01585 mol\text{moles} = M \times V = 0.1585\ \text{mol/L} \times 0.1000\ \text{L} = 0.01585\ \text{mol}
Step 3: Convert moles to mass.
Molar mass of NaH₂PO₄ ≈ 119.98 g/mol mass=0.01585 mol×119.98 g/mol=1.902 g\text{mass} = 0.01585\ \text{mol} \times 119.98\ \text{g/mol} = 1.902\ \text{g}
Correction: That value seems too high. Let’s recheck the calculation.
Actually, we forgot to subtract the existing base amount when interpreting the ratio. Since we know the base is 0.1 M and the ratio is: baseacid=10(7.00−7.20)=0.631\frac{\text{base}}{\text{acid}} = 10^{(7.00 – 7.20)} = 0.631
Then: 0.100acid=0.631⇒acid=0.1000.631=0.1585 mol/L\frac{0.100}{\text{acid}} = 0.631 \Rightarrow \text{acid} = \frac{0.100}{0.631} = 0.1585\ \text{mol/L}
That’s valid.
Now moles of acid needed: 0.1585 mol/L×0.1 L=0.01585 mol0.1585\ \text{mol/L} \times 0.1\ \text{L} = 0.01585\ \text{mol}
Mass: 0.01585×119.98=1.902 g(final and correct)0.01585 \times 119.98 = \boxed{1.902\ \text{g}} \quad \text{(final and correct)}
Apologies for earlier confusion. The correct mass of NaH₂PO₄ to add is 1.902 grams.