Calculate the amount of sodium dihydrogen phosphate, NaH2PO4 (molecular weight = 120 g/mol) and hydrated sodium hydrogen phosphate, Na2HPO4 (molecular weight = 142 g/mol) needed to make 50 mL of 20 mM pH 7.2 buffer. (pKa of the buffer = 7.2).
The Correct Answer and Explanation is:
To calculate the amount of sodium dihydrogen phosphate (NaH₂PO₄) and sodium hydrogen phosphate (Na₂HPO₄) required to prepare 50 mL of 20 mM phosphate buffer at pH 7.2, we will use the Henderson-Hasselbalch equation and basic stoichiometry.
Step 1: Understand 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])
Here:
- pH = 7.2
- pKa = 7.2 (given)
- Base = Na₂HPO₄ (conjugate base, HPO₄²⁻)
- Acid = NaH₂PO₄ (weak acid, H₂PO₄⁻)
Substituting:7.2=7.2+log([base][acid])⇒log([base][acid])=0⇒[base][acid]=17.2 = 7.2 + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right) \Rightarrow \log\left(\frac{[\text{base}]}{[\text{acid}]}\right) = 0 \Rightarrow \frac{[\text{base}]}{[\text{acid}]} = 17.2=7.2+log([acid][base])⇒log([acid][base])=0⇒[acid][base]=1
So, equal concentrations of base and acid are needed.
Step 2: Total Buffer Concentration
We want a total concentration of 20 mM for the buffer in 50 mL.
Since base and acid concentrations are equal:[base]=[acid]=20 mM2=10 mM[\text{base}] = [\text{acid}] = \frac{20\ \text{mM}}{2} = 10\ \text{mM}[base]=[acid]=220 mM=10 mM
Step 3: Calculate Moles Needed
Volume = 50 mL = 0.050 L
Moles of each component:moles of base (Na₂HPO₄)=10×10−3 mol/L×0.050 L=5.0×10−4 mol\text{moles of base (Na₂HPO₄)} = 10 \times 10^{-3}\ \text{mol/L} \times 0.050\ \text{L} = 5.0 \times 10^{-4}\ \text{mol}moles of base (Na₂HPO₄)=10×10−3 mol/L×0.050 L=5.0×10−4 molmoles of acid (NaH₂PO₄)=5.0×10−4 mol\text{moles of acid (NaH₂PO₄)} = 5.0 \times 10^{-4}\ \text{mol}moles of acid (NaH₂PO₄)=5.0×10−4 mol
Step 4: Convert Moles to Grams
Molar mass:
- Na₂HPO₄ = 142 g/mol
- NaH₂PO₄ = 120 g/mol
mass of Na₂HPO₄=5.0×10−4×142=0.071 g\text{mass of Na₂HPO₄} = 5.0 \times 10^{-4} \times 142 = 0.071\ \text{g}mass of Na₂HPO₄=5.0×10−4×142=0.071 gmass of NaH₂PO₄=5.0×10−4×120=0.060 g\text{mass of NaH₂PO₄} = 5.0 \times 10^{-4} \times 120 = 0.060\ \text{g}mass of NaH₂PO₄=5.0×10−4×120=0.060 g
Final Answer:
- 0.060 g of NaH₂PO₄
- 0.071 g of Na₂HPO₄
Explanation
Buffers are solutions that resist changes in pH when small amounts of acid or base are added. A phosphate buffer commonly uses sodium dihydrogen phosphate (NaH₂PO₄) and disodium hydrogen phosphate (Na₂HPO₄) as the weak acid and its conjugate base. The pKa for this buffer system is 7.2, which is ideal for maintaining a near-neutral pH.
To prepare a buffer of a specific pH, the Henderson-Hasselbalch equation is used. This equation relates the pH of the buffer to the ratio of base to acid. If the desired pH is equal to the pKa, then the ratio of base to acid becomes 1. This means equal amounts, in terms of concentration, of the weak acid and its conjugate base must be present.
In this question, the target is a 20 mM phosphate buffer in a total volume of 50 mL. Since the base and acid concentrations are equal, each must contribute 10 mM. To find the mass of each compound required, we calculate the number of moles using the molarity formula and then convert to grams using the molar mass of each salt.
For 10 mM in 50 mL, this equates to 0.0005 mol of each compound. Multiplying these mole values by their respective molar masses gives the required mass. The results show that 0.060 g of sodium dihydrogen phosphate and 0.071 g of disodium hydrogen phosphate are needed to make the buffer solution. This precise balance will ensure that the solution maintains a stable pH of 7.2.
