A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of
\(\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH] }\)
required to make buffer is _____.
Given : Ka(CH3CH2COOH) = 1.3 × 10–5
0.03
0.13
0.23
0.33
The correct answer is (B) : 0.13
\(CH_3CH_2COOH⇌CH_3CH_2COO^−+H^+\)
From Henderson equation
\(pH=pK_a+log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}\)
\(4=−log 1.3×10^{−5} log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}\)
\(−log^{10^{−4}}=−log1.3×10^{−5}+log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}\)
\(−log^{10^{−4}}=−log1.3×10^{−5} \frac{[CH_3CH_2COOH]}{[CH_3CH_2COO^-]}\)
\(10^{−4}=1.3×10^{−5}\frac{[CH_3CH_2COOH]}{[CH_3CH_2COO^-]}\)
\(\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}=0.13\)
\[ 2X(g) \rightleftharpoons 2Y(g) + Z(g) \]
$K_C$ at 400K is $1 \times 10^{-3}$ mol L-1. What is the value of $K_P$ for the equilibrium at 400K?\[ R = 0.082 \, \text{L atm K}^{-1} \, \text{mol}^{-1} \]
The value of current \( I \) in the electrical circuit as given below, when the potential at \( A \) is equal to the potential at \( B \), will be _____ A.
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.
Law of Chemical Equilibrium states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to a power equal to the corresponding stoichiometric coefficients as represented by the balanced chemical equation.
Let us consider a general reversible reaction;
A+B ↔ C+D
After some time, there is a reduction in reactants A and B and an accumulation of the products C and D. As a result, the rate of the forward reaction decreases and that of backward reaction increases.
Eventually, the two reactions occur at the same rate and a state of equilibrium is attained.
By applying the Law of Mass Action;
The rate of forward reaction;
Rf = Kf [A]a [B]b
The rate of backward reaction;
Rb = Kb [C]c [D]d
Where,
[A], [B], [C] and [D] are the concentrations of A, B, C and D at equilibrium respectively.
a, b, c, and d are the stoichiometric coefficients of A, B, C and D respectively.
Kf and Kb are the rate constants of forward and backward reactions.
However, at equilibrium,
Rate of forward reaction = Rate of backward reaction.
Kc is called the equilibrium constant expressed in terms of molar concentrations.
The above equation is known as the equation of Law of Chemical Equilibrium.