MPBSE Class 11 Chemistry Solutions For Equilibrium

Equilibrium Class 11 Chemistry Notes

Question 1. HPO₄^2- can act both as a Bronsted base and as a Bronsted acid. Write the equation of equilibrium established by HPO₄^2- as an acid and a base in an aqueous solution. Also, write the expressions of K_a & K_b in two cases.  What are the conjugate acid and base of HS^–?
Answer:

HPO₄^2-  can denate and accept protons in aqueous solution. Thus it can serve both as an acid and base.

⇒ \(\begin{aligned}
& \mathrm{HPO}_4^{2-}(a q)[\text { Acid }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{PO}_4^{3-}(a q)+\mathrm{H}_3 \mathrm{O}_2^{+}(a q) \\
& \mathrm{HPO}_4^{2-}(a q)[\text { Base }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_2 \mathrm{PO}_4^{-}(a q)+\mathrm{OH}^{-}(a q)
\end{aligned}\)

⇒ \(K_a=\frac{\left[\mathrm{PO}_4^{3-}\right] \times\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{HPO}_4^{2-}\right]} ; K_b=\frac{\left[\mathrm{H}_2 \mathrm{PO}_4^{-}\right] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{HPO}_4^{2-}\right]}\)

HS^– (aq) + H₂O(l) ⇌  H₂S(aq) + OH^– (aq)

Hence, the conjugate acid of HS^–  is H₂S.

HS^–(aq) + H₂O(l) ⇌  S^2-(aq) + H₃O+(aq)

Therefore, the conjugate base of HS^– is S^2--.

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Question 2. An aqueous solution of sodium bisulfate is acidic, whereas an aqueous solution of sodium bicarbonate is basic—Explain.
Answer:

Since sodium bisulfate (NaHSO₄) is a salt of strong acid (H₂SO₄) and strong base (NaOH), it is not hydrolyzed in an aqueous solution. NaHS04 in its solution dissociates completely to form Na⁺ and HSO₄ ions and HSO₄ ions so formed get ionized to form H₃O+ and SO₂^-4 ions. As a result, the aqueous solution of NaHSO₄ becomes acidic.

⇒ \(\begin{gathered}
\mathrm{NaHSO}_4(a q) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{HSO}_4^{-}(a q) \\
\mathrm{HSO}_4^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{SO}_4^{2-}(a q)
\end{gathered}\)

NaHCO₃ in its solution dissociates completely to form Na+ and HCO₃ ions [NaHCO₃(aq)-Na+(aq) + HCO₃ (aq) ]. HCO₃ can act both as an acid and a base in aqueous solution.

⇒ \(\begin{aligned}
& \mathrm{HCO}_3^{-}(a q)[\text { Acid }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{CO}_3^{2-}(a q) \\
& \mathrm{HCO}_3^{-}(a q)[\text { Base }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3(a q)+\mathrm{OH}^{-}(a q)
\end{aligned}\)

Equilibrium Class 11 Chemistry Notes

Question 3. Calculate the formation constant of [Ag(NH₃)₂]⁺
Answer:

⇒  \(\mathrm{Ag}^{+}(a q)+\mathrm{NH}_3(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} ; K_1=3.5 \times 10^3\)

⇒ \(\begin{aligned}
{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} ; } \\
K_1=1.7 \times 10^3 \ldots
\end{aligned}\)

Equation (3) represents the formation reaction of [Ag(NH₃)₂]2+. Therefore, the equilibrium constant for the reaction represented by equation (3) will be the formation constant for [Ag(NH₃)₂]2+. As equation (3) is obtained by adding equations (1) and (2), the formation constant (fcy) for [Ag(NH₃)₂]2+ equals ky x k2.

Thus, k_y = k_y X k₂ = (3.5 × 10³) × (1.7 ×  10³) = 5.95 ×  10⁶

Question 4. The first and second dissociation constants of an acid H₂A are 1 × 10^-5 and 5 × 10^-10  respectively. Calculate the value of the overall dissociation constant.
Answer:

⇒ \(\begin{aligned}
\mathrm{H}_2 \mathrm{~A}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{HA}^{-}(a q) ; \\
K_1=1 \times 10^{-5} \ldots(1)
\end{aligned}\)

⇒ \(\begin{array}{r}
\mathrm{HA}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{A}^{2-}(a q) ; \\
K_2=5 \times 10^{-10} \ldots
\end{array}\)

The overall dissociation reaction is the sum of the. reactions (1) and (2). H₂A(aq) + 2H₂O(Z) 2H30+(aq) + A3-(aq) Thus, the overall disputation constant K = K₁ × K₂ =  1 × 10^-5 and 5 × 10^-10= 5 ×10^-15

Question 5. a -D-glucose Beta -D glucose, the equilibrium constant for this is 1.8. Calculate the percentage of a -D glucose at equilibrium.
Answer:

Suppose, the initial concentration of a -D glucose is a mol L^-1 and its degree of conversion to p -D glucose at equilibrium is x mol-L^-1

Chemical Equilibrium Class 11 Questions

Therefore, the concentration of a -D glucose and D glucose at equilibrium will be as follows—

⇒ \(\begin{array}{ccc}
\begin{array}{c}
\text { Initial concentration } \\
\left(\text { in mol } \cdot \mathrm{L}^{-1}\right. \text { ) }
\end{array} & a & 0 \\
\begin{array}{c}
\text { Equilibrium concentration } \\
\left(\text { in mol } \cdot \mathrm{L}^{-1}\right. \text { ) }
\end{array} & a-a x & a x
\end{array}\)

Equilibrium constant for this process, \(K=\frac{[\beta-\mathrm{D}-\text { glucose }]}{[\alpha-\mathrm{D}-\text { glucose }]}\)

⇒ \(1.8=\frac{a x}{a(1-x)}=\frac{x}{1-x}\) or, \(x=\frac{1.8}{2.8}=0.6428\)

Thus, the percentage of a -D-glucose at equilibrium is \(\frac{a(1-0.6428)}{a} \times 100=35.72 \%\)

Question 6. Solid Ba(NO₃)₂ is gradually dissolved in a 1.0 x 10-4 M Na₂CO₃ solution. At what concentration of Ba2+ will a precipitate begin to form? ( Ksp for BaCO₃ = 5.1 × 10^-9 )
Answer:

Ba(NO₃)₂ reacts with Na₂CO₃ to form BaCO₃. BaCO₃ is a sparingly soluble compound that forms the following equilibrium in its saturated solution.

⇒ \(\mathrm{BaCO}_3(s) \rightleftharpoons \mathrm{Ba}^{2+}(a q)+\mathrm{CO}^{2-}(a q)\)

For \(\mathrm{BaCO}_3, K_{s p}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{CO}_3^{2-}\right]\)

In the solution \(\left[\mathrm{CO}_3^{2-}\right]=1 \times 10^{-4} \mathrm{M}\)

BaCO₃ will precipitate when [Ba²⁺][CO₂^–] → Ksp, i.e., when

⇒ \(\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{CO}_3^{2-}\right]>5.1 \times 10^{-9}\left(\text { as } K_{s p}\left[\mathrm{BaCO}_3\right]=5.1 \times 10^{-9}\right)\) \(\text { If }\left[\mathrm{CO}_3^{2-}\right]=1 \times 10^{-4} \mathrm{M} \text {, then }\left[\mathrm{Ba}^{2+}\right]>\frac{5.1 \times 10^{-9}}{1 \times 10^{-4}}\)

Or, [Ba²⁺] > 5.1 × 10^-5M

Thus, the precipitation of BaCO₃ will start when [Ba2+] in the solution is grater than 5.1 × 10^-5 M.

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Question 7. 2.5 m l of \(\frac{2}{5} \mathrm{M}\) weak monoacdic base Kb = 1 x 10-125at 25°C ) is titrated with \(\frac{2}{15}\) in water at 25°C. Calculate the concentration of H + at the equivalence point. (kw = 1 × 10-14)
Answer:

Ana. 2.5 mL of \(\frac{2}{15}\) M monobasic acid \(\equiv \frac{2}{5} \times 2.5 \equiv 1\) mmol of the base.

Suppose, VmL of \(\frac{2}{15} \mathrm{M}\) HC1 is required for the neutralisation. \(V \mathrm{~mL} \text { of } \frac{2}{15} \mathrm{M} \mathrm{HCl} \equiv \frac{2}{15} \times V \mathrm{mmol} \mathrm{HCl} .\)

Therefore \(\) mmol HCL will neutralise 1mmol of the base Hence \(\frac{2 \times V}{15}=1 \text { or, } V=7.5 \mathrm{~mL}\)

The total volume of the solution after neutralization = (2.5 + 7.5) mL = 10 mL.

The number of mmol of the salt formed in the neutralization =1 mmol.

So, the concentration of the salt in the final solution \((C)=\frac{1}{10}=0.1 \mathrm{M}\) As the resulting salt is formed from a weak base and a strong acid, it undergoes hydrolysis. The pH of the solution of such a salt is given by the pH
\(=7-\frac{1}{2} p K_b-\frac{1}{2} \log C\) \(\text { As } K_b=10^{-12}, p K_b=-\log _{10}\left(10^{-12}\right)=12\)

Therefore \(p H=7-\frac{1}{2} \times 12-\frac{1}{2} \log (0.1)=7-6+0.5=1.5\) and the concentration of H+(aq) ions at the equivalence point is [H⁺]

= 10-PH M = 10-1-5 M = 0.316 M.