Respuesta :
Answer:
[tex]Mass_{sodium\ formate}= 889.57\ g[/tex]
Explanation:
Considering the Henderson- Hasselbalch equation for the calculation of the pH of the acidic buffer solution as:
[tex]pH=pK_a+log\frac{[salt]}{[acid]}[/tex]
Given that:-
[Acid] = 0.12 M
Volume = 3.0 L
pKa = 3.74
pH = 5.30
So,
[tex]5.30=3.74+log\frac{[sodium\ formate]}{0.12}[/tex]
Solving, we get that:-
[Sodium formate] = 4.36 M
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
So,
[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]
So, Moles of sodium formate = 4.36*3.0 moles = 13.08 moles
Molar mass of sodium formate = 68.01 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]13.08\ mole= \frac{Mass}{68.01\ g/mol}[/tex]
[tex]Mass_{sodium\ formate}= 889.57\ g[/tex]
