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
\(pH=pK_a+log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}\)
\(4=−log 1.3×10^{−5} log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}\)
\(−log^{10^{−4}}=−log1.3×10^{−5}+log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}\)
\(−log^{10^{−4}}=−log1.3×10^{−5} \frac{[CH_3CH_2COOH]}{[CH_3CH_2COO^-]}\)
\(10^{−4}=1.3×10^{−5}\frac{[CH_3CH_2COOH]}{[CH_3CH_2COO^-]}\)
\(\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}=0.13\)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
At \(-20^\circ \text{C}\) and 1 atm pressure, a cylinder is filled with an equal number of \(H_2\), \(I_2\), and \(HI\) molecules for the reaction:
\[H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\] The \(K_P\) for the process is \(x \times 10^{-1}\).
(x = ___________)
Given: \(R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1}\)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
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.