Consider a weak base 'B' of $$pK_{b}=5.699 $$. 'x' mL of 0.02 M HCI and 'y' mL of 0.02 M weak base 'B' are mixed to make 100 mL of a buffer of pH 9 at 25 °C. The values of 'x' and 'y' respectively are:
[Given: log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.699]

We need to find the values of $$x$$ and $$y$$ to prepare a 100 mL basic buffer of pH 9 from a weak base $$B$$ ($$pK_b = 5.699$$) and strong acid $$\text{HCl}$$.

1. Buffer Equations:

• Calculate $$\text{pOH}$$:
  $$\text{pOH} = 14 - \text{pH} = 14 - 9 = 5$$
• Henderson-Hasselbalch Equation for Basic Buffer:
  $$\text{pOH} = pK_b + \log\left(\frac{[\text{conjugate acid}]}{[\text{weak base}]}\right)$$
  $$5 = 5.699 + \log\left(\frac{[\text{BH}^+]}{[B]}\right)$$
  $$\log\left(\frac{[\text{BH}^+]}{[B]}\right) = -0.699$$
  $$\frac{[\text{BH}^+]}{[B]} = 10^{-0.699} = 10^{\log 2 - 1} = 0.2 = \frac{1}{5}$$

2. Reaction and Mole Balance:

When $$x\text{ mL}$$ of $$0.02\text{ M HCl}$$ reacts with $$y\text{ mL}$$ of $$0.02\text{ M B}$$:

• $$\text{Initial millimoles of HCl} = 0.02x$$
• $$\text{Initial millimoles of B} = 0.02y$$
• Since $$\text{HCl}$$ is the limiting reagent, it reacts completely to form conjugate acid $$\text{BH}^+$$:
  • $$\text{Millimoles of conjugate acid } (\text{BH}^+) = 0.02x$$
  • $$\text{Millimoles of remaining weak base } (B) = 0.02y - 0.02x$$
• Substitute into the ratio:
  $$\frac{0.02x}{0.02y - 0.02x} = \frac{1}{5} \implies \frac{x}{y - x} = \frac{1}{5}$$
  $$5x = y - x \implies y = 6x$$
• Using total volume:
  $$x + y = 100\text{ mL}$$
  $$x + 6x = 100 \implies 7x = 100 \implies x = 14.3\text{ mL}$$
  $$y = 100 - 14.3 = 85.7\text{ mL}$$

Conclusion:

The required volumes of $$\text{HCl}$$ and weak base $$B$$ are $$14.3\text{ mL}$$ and $$85.7\text{ mL}$$, respectively.

Answer: Option B — $$x = 14.3, y = 85.7$$