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
\(pH=pK_a+log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}\)
\(4=−log 1.3×10^{−5} log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}\)
\(−log^{10^{−4}}=−log1.3×10^{−5}+log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}\)
\(−log^{10^{−4}}=−log1.3×10^{−5} \frac{[CH_3CH_2COOH]}{[CH_3CH_2COO^-]}\)
\(10^{−4}=1.3×10^{−5}\frac{[CH_3CH_2COOH]}{[CH_3CH_2COO^-]}\)
\(\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}=0.13\)
For the reaction A(g) $\rightleftharpoons$ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given: R = 0.0831 atm $mol^{–1} K^{–1}$]
$K_p$ for the reaction at 1000 K is:
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is:
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.