Solution: To find the volume of benzoic acid in the buffer solution, we can use the Henderson-Hasselbalch equation:
\[ \text{pH} = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) \]
Where: \([A^-]\) is the concentration of the conjugate base (sodium benzoate) and \([HA]\) is the concentration of the weak acid (benzoic acid).
Given Information: pH = 4.5
\[ pK_a = 4.20, \quad [A^-] = 1 \, M \text{ (sodium benzoate)} \]
Total volume of buffer solution = 300 mL.
Calculate the ratio of base to acid:
\[ 4.5 = 4.20 + \log \left( \frac{[A^-]}{[HA]} \right) \]
\[ 0.30 = \log \left( \frac{1}{[HA]} \right) \]
\[ 10^{0.30} = \frac{1}{[HA]} \implies [HA] = \frac{1}{10^{0.30}} \approx 0.50 \, M \]
Calculate the volume of benzoic acid: Let \(V_a\) be the volume of 1M benzoic acid. The concentration in the total 300 mL buffer solution:
\[ [HA] = \frac{V_a}{200} \]
Setting the concentrations equal gives:
\[ 0.50 = \frac{V_a}{200} \implies V_a = 0.50 \times 200 = 100 \, mL \]
Total volume of benzoic acid solution: To find the total volume of benzoic acid needed to maintain the desired pH:
\[ V_a = 100 \, mL \quad \text{(rounded from calculations)} \]
Thus, the volume of benzoic acid solution in 300 mL of the buffer solution is: 100 mL
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to: