Answer: The mass of base required is 3.87 g.
Explanation:
The given data is as follows.
pH of buffer = 2.0, volume of the solution (V) = 100 mL,
molarity of HA, [HA] = 0.1 M, [tex]K_{a} = 1.20 \times 10^{-2}[/tex]
Now, we will calculate the value of [tex]pK_{a}[/tex] as follows.
[tex]pK_{a} = -log (K_{a})[/tex]
= [tex]-log (1.20 \times 10^{-2})[/tex]
= 1.92
According to Henderson equation,
pH = [tex]pK_{a} + log \frac{[A^{-}]}{[HA]}[/tex]
2.0 = [tex]1.92 + log \frac{[A^{-}]}{0.1 M}[/tex]
[tex][A^{-}][/tex] = 0.12 M
Moles of base = [tex]0.12 M \times 100 mL \times \frac{1 L}{1000 mL}[/tex]
= 0.012 mol
Hence, mass of the base will be calculated as follows.
[tex]0.012 mol \times \frac{322.1 g}{1 mol Na_{2}SO_{4}.10H_{2}O}[/tex]
= 3.87 g
Hence, the mass of base required is 3.87 g.
The required mass of solid conjugate base will be:
"3.87 g".
According to the question,
pH of buffer = 2.00
Volume of solution, V = 100.00 mL
Molarity of HA, [HA] = 0.1 M
The value of p[tex]K_a[/tex] will be:
= -log([tex]K_a[/tex])
= -log(1.20 × 10⁻²)
= 1.92
By using Henderson equation,
→ pH = p[tex]K_a[/tex] + log [tex]\frac{[A^-]}{HA}[/tex]
By substituting the values,
2.0 = 1.92 + log [tex]\frac{[A^-]}{0.1}[/tex]
[A⁻] = 0.12 M
Now,
Moles of base will be:
= 0.12 × 100 × [tex]\frac{1 \ L}{1000 \ mL}[/tex]
= 0.012 mol
Hence,
The required base mass be:
= 0.012 × [tex]\frac{322.1}{1 \ mol \ Na_2SO_4. 10 H_2O}[/tex]
= 3.87 g
Thus the above answer is correct.
Find out more information about buffer solution here:
https://brainly.com/question/22390063
