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
\(pH=pK_a+log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}\)
\(4=−log 1.3×10^{−5} log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}\)
\(−log^{10^{−4}}=−log1.3×10^{−5}+log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}\)
\(−log^{10^{−4}}=−log1.3×10^{−5} \frac{[CH_3CH_2COOH]}{[CH_3CH_2COO^-]}\)
\(10^{−4}=1.3×10^{−5}\frac{[CH_3CH_2COOH]}{[CH_3CH_2COO^-]}\)
\(\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}=0.13\)
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).

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.