Answer :
To find the pH of the solution formed by mixing 100 mL of 0.2 M HCOOH (formic acid) with 150 mL of 0.4 M HCOONa (sodium formate), we use the concept of a buffer solution. A buffer solution is formed when a weak acid and its conjugate base are mixed.
Step-by-step solution:
Determine the moles of HCOOH and HCOONa:
- Moles of HCOOH = [tex]0.2 \, \text{M} \times 0.1 \, \text{L} = 0.02 \, \text{moles}[/tex]
- Moles of HCOONa = [tex]0.4 \, \text{M} \times 0.15 \, \text{L} = 0.06 \, \text{moles}[/tex]
Identify the buffer components and their concentrations:
The buffer is made of HCOOH (weak acid) and HCOONa (its conjugate base, which supplies HCOO⁻). Since the volume changes upon mixing (100 mL + 150 mL = 250 mL = 0.25 L), the concentrations change too.
- Concentration of HCOOH = [tex]\frac{0.02 \, \text{moles}}{0.25 \, \text{L}} = 0.08 \, \text{M}[/tex]
- Concentration of HCOO⁻ = [tex]\frac{0.06 \, \text{moles}}{0.25 \, \text{L}} = 0.24 \, \text{M}[/tex]
Use the Henderson-Hasselbalch equation to find pH:
The Henderson-Hasselbalch equation for a buffer is:
[tex]\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]\right)[/tex]
Given [tex]\text{K}_a = 1.0 \times 10^{-4}[/tex], so [tex]\text{pK}_a = -\log(1.0 \times 10^{-4}) = 4[/tex].
Plug these values into the equation:
[tex]\text{pH} = 4 + \log\left(\frac{0.24}{0.08}\right) = 4 + \log(3)[/tex]
[tex]\log(3) \approx 0.4771\text{, therefore :}[/tex]
[tex]\text{pH} \approx 4 + 0.4771 = 4.4771[/tex]
Find the closes option:
The pH of the solution is closest to [tex]5 - \log(3.3)[/tex] as [tex]\log(3.3)[/tex] approximates [tex]0.5051[/tex], so [tex]5 - 0.5051 = 4.4949[/tex].
Therefore, the answer is option (a): 5 - log3,3.