Answer:
0.387 g
Explanation:
pH of the buffer = 1
V = Volume of solution = 100 mL
[HA] = Molarity of HA = 0.1 M
[tex]K_a[/tex] = Acid dissociation constant = [tex]1.2\times 10^{-2}[/tex]
(assuming base as [tex]Na_2SO_410H_2O[/tex])
Molar mass of base = 322.2 g/mol
pKa is given by
[tex]pK_a=-\log K_a\\\Rightarrow pKa=-\log(1.2\times 10^{-2})\\\Rightarrow pK_a=1.92[/tex]
From the Henderson-Hasselbalch equation we get
[tex]pH=pK_a+\log\dfrac{[A^-]}{[HA]}\\\Rightarrow pH-pK_a=\log\dfrac{[A^-]}{[HA]}\\\Rightarrow 10^{pH-pK_a}=\dfrac{[A^-]}{[HA]}\\\Rightarrow [A^-]=10^{pH-pK_a}[HA]\\\Rightarrow [A^-]=10^{1-1.92}\times0.1\\\Rightarrow [A^-]=0.01202\ \text{M}[/tex]
Moles of base
[tex]0.01202\times100\times\dfrac{1}{10^3}=0.001202\ \text{moles}[/tex]
Mass of base is given by
[tex]0.001202\times 322.2=0.387\ \text{g}[/tex]
The required mass of the base is 0.387 g.
