The pKa of formic acid (HCOOH) is 3.74. A buffer contains a mixture of formic acid and sodium formate (NaHCOO). The pH of the buffer solution is 3.74. Which statement below accurately depicts the relative amounts of HCOOH and NaHCOO in the buffer? (Hint: Consider the Henderson-Hasselbalch equation.) [HCOOH] = [NaHCOO] [HCOOH] > [NaHCOO] [HCOOH] << [NaHCOO] It is not possible to make a buffer at this pH from these compounds. [HCOOH] < [NaHCOO]
The Correct Answer and Explanation is:
Correct Answer: [HCOOH] = [NaHCOO]
Explanation:
To determine the relative concentrations of formic acid (HCOOH) and sodium formate (NaHCOO) in the buffer solution, we use the Henderson-Hasselbalch equation, which relates the pH of a buffer to the pKa of the weak acid and the ratio of the conjugate base to the acid:pH=pKa+log([A−][HA])\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)pH=pKa+log([HA][A−])
Where:
- pH is the acidity of the buffer solution
- pKa is the acid dissociation constant of the weak acid
- [A⁻] is the concentration of the conjugate base (in this case, NaHCOO, the sodium salt of formic acid)
- [HA] is the concentration of the weak acid (HCOOH, formic acid)
Given:
- pKa of formic acid = 3.74
- pH of the buffer = 3.74
Now substitute into the Henderson-Hasselbalch equation:3.74=3.74+log([NaHCOO][HCOOH])3.74 = 3.74 + \log\left(\frac{[\text{NaHCOO}]}{[\text{HCOOH}]}\right)3.74=3.74+log([HCOOH][NaHCOO])
Subtract 3.74 from both sides:0=log([NaHCOO][HCOOH])0 = \log\left(\frac{[\text{NaHCOO}]}{[\text{HCOOH}]}\right)0=log([HCOOH][NaHCOO])
This means:[NaHCOO][HCOOH]=100=1\frac{[\text{NaHCOO}]}{[\text{HCOOH}]} = 10^0 = 1[HCOOH][NaHCOO]=100=1
Therefore, the concentration of sodium formate (the conjugate base) is equal to the concentration of formic acid (the weak acid). When the pH equals the pKa, this is always the case. This is a characteristic feature of buffers—they are most effective when pH = pKa, as they can neutralize both added acids and bases equally well.
Conclusion:
Since pH = pKa, [HCOOH] = [NaHCOO]. This balance makes the buffer highly effective at resisting changes in pH.
