The pKa of sodium dihydrogen phosphate (NaH2PO4) is 6.8. What ratio of hydrogen phosphate (HPO4^2-) to dihydrogen phosphate (H2PO4^-) would create a buffer of pH 7.5? The acid dissociation reaction of dihydrogen phosphate into hydrogen phosphate is shown below: H2PO4^- + H2O ⇌ HPO4^2- + H3O+. pKa = 6.8
The Correct Answer and Explanation is:
To determine the ratio of hydrogen phosphate (HPO₄²⁻) to dihydrogen phosphate (H₂PO₄⁻) needed to create a buffer of pH 7.5, we use the Henderson-Hasselbalch equation:pH=pKa+log([base][acid])\text{pH} = \text{pKa} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)pH=pKa+log([acid][base])
In this reaction:
- The acid is dihydrogen phosphate (H₂PO₄⁻)
- The base is hydrogen phosphate (HPO₄²⁻)
Given:
- pKa = 6.8
- Desired pH = 7.5
Step 1: Plug values into the equation
7.5=6.8+log([HPO₄2−][H₂PO₄−])7.5 = 6.8 + \log \left( \frac{[\text{HPO₄}^{2-}]}{[\text{H₂PO₄}^-]} \right)7.5=6.8+log([H₂PO₄−][HPO₄2−])
Step 2: Subtract pKa from both sides
7.5−6.8=log([HPO₄2−][H₂PO₄−])7.5 – 6.8 = \log \left( \frac{[\text{HPO₄}^{2-}]}{[\text{H₂PO₄}^-]} \right)7.5−6.8=log([H₂PO₄−][HPO₄2−])0.7=log([HPO₄2−][H₂PO₄−])0.7 = \log \left( \frac{[\text{HPO₄}^{2-}]}{[\text{H₂PO₄}^-]} \right)0.7=log([H₂PO₄−][HPO₄2−])
Step 3: Take the antilogarithm (base 10)
[HPO₄2−][H₂PO₄−]=100.7≈5.01\frac{[\text{HPO₄}^{2-}]}{[\text{H₂PO₄}^-]} = 10^{0.7} \approx 5.01[H₂PO₄−][HPO₄2−]=100.7≈5.01
Final Answer:
The ratio of HPO₄²⁻ to H₂PO₄⁻ must be approximately 5.0 to 1 in order to prepare a buffer solution at pH 7.5 using sodium dihydrogen phosphate.
Explanation (around 300 words):
A buffer solution is a mixture of a weak acid and its conjugate base, designed to resist changes in pH. The effectiveness of a buffer depends on the ratio of the concentrations of the base to the acid, and this is described using the Henderson-Hasselbalch equation.
In this case, we are given that sodium dihydrogen phosphate (NaH₂PO₄) has a pKa of 6.8. The conjugate base of this acid is hydrogen phosphate (HPO₄²⁻). We want to create a buffer with a pH of 7.5. Since 7.5 is higher than 6.8, we will need more base than acid in our solution.
By rearranging and solving the Henderson-Hasselbalch equation, we find that the logarithmic difference between the desired pH and the pKa is 0.7. Taking the antilog (10 raised to the power of 0.7), we calculate the necessary ratio of base to acid to be about 5.0. This means we need five times as much hydrogen phosphate as dihydrogen phosphate to reach a pH of 7.5.
This ratio allows the buffer to maintain its pH when small amounts of acid or base are added, which is useful in biological and chemical systems where stable pH is essential.
