Answer:
pH = 5.80
Explanation:
The buffer solution is:
H₂CO₃(aq) + H₂O(l) ⇄ HCO₃⁻Na⁺(aq) + H₃O⁺(aq)
To find the pH of the buffer solution we will use the Henderson-Hasselbalch equation:
[tex]pH = pKa + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})[/tex] (1)
First, we need to find the concentration of the buffer solution. For the NaHCO₃ we have:
[tex][NaHCO_{3}] = \frac{mol}{V} = \frac{m}{M*V}[/tex]
Where:
m: is the mass of the NaHCO₃ = 7.20 g
M: is the molar mass of the NaHCO₃ = 84.007 g/mol
V: is the volume of the solution = 400.0 mL
Hence, the concentration of NaHCO₃ is:
[tex][NaHCO_{3}] = \frac{7.20 g}{84.007 g/mol*400.0 \cdot 10^{-3} L} = 0.214 M[/tex]
Now, the concentration of H₂CO₃ is:
[tex] V_{i}C_{i} = V_{f}C_{f} [/tex]
Where:
Vi: is the initial volume of H₂CO₃ = 56.0 mL
Ci: is the initial concentration of H₂CO₃ = 5.60 M
Vf: is the final volume of H₂CO₃ = 400.0 mL
Cf: is the final concentration of H₂CO₃ (to find)
[tex] C_{f} = \frac{V_{i}C_{i}}{V_{f}} = \frac{56.0 mL*5.60 M}{400.0 mL} = 0.784 M [/tex]
Finally, we can use the equation (1) to find the pH of the buffer solution:
[tex] pH = -log(4.3 \cdot 10^{-7}) + log(\frac{0.214 M}{0.784 M}) = 5.80 [/tex]
I hope it helps you!
