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Equilibrium Test - 72

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Equilibrium Test - 72
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  • Question 1
    1 / -0
    Solubility of  a solute in a solvent (say water) is dependent on temperature as given by $$ S = A.e^{- \Delta HlRT} $$ , where $$ \Delta H$$ is heat of reaction : solute $$ + H_2O \leftrightharpoons $$ solution. For a given solution, a variation of log S with temperature is shown graphically the solute should be:

    Solution

  • Question 2
    1 / -0
    Calculate the formation constant for the reaction of a tripositive metal ion with thiocyanate ions to form the monocomplex if the total metal concentration in the solution is $$2 \times 10^{-3} M$$, the total $$SCN^{-}$$ concentration is $$1.51 \times 10^{-3} M$$ and the free $$SCN^{-}$$ concentration is $$1.0 \times 10^{-5} M$$.
    Solution

  • Question 3
    1 / -0
    A buffer solution is $$0.25 M - CH_{3}COOH + 0.15 M - CH_{3}COONa$$, saturated in $$H_{2}S (0.1 M)$$ and has $$[Mn^{2+}] = 0.04 M$$, $$K_{a}(CH_{3}COOH) = 2.0 \times 10^{-5}$$, $$ K_{a} (H_{2}S) = 1.0 \times 10^{-21}$$ and $$K_{sp}(MnS) = 2.5 \times 10^{-13}$$.
    Which buffer component should be increased in concentration and to which minimum value to just start precipitation of $$MnS$$?
  • Question 4
    1 / -0
    An amount of $$0.10$$ moles of $$AgCl(s)$$ is added to one litre of water. Next, the crystals of $$NaBr$$ are added until $$75$$% of the $$AgCl$$ is converted to $$AgBr(s)$$, the less soluble silver halide. What is $$Br^{-}$$ at this point? 

    $$K_{sp}$$ of $$AgCl = 2 \times 10^{-10}$$ and $$K_{sp}$$ of $$AgBr = 4 \times 10^{-13}$$.
    Solution

  • Question 5
    1 / -0
    An amount of $$0.01$$ moles of solid $$AgCN$$ is rendered soluble in $$1 l$$ by adding just sufficient excess cyanide ion to form $$Ag(CN)^{2-}$$ and the concentration of free cyanide ion is $$2.5 \times 10^{-7} M$$. Determine $$[Ag^{+}]$$ in the solution neglecting hydrolysis of cyanide ion. $$K_{diss}$$ for $$Ag(CN)_{2}^{-} = 1.0 \times 10^{-20}$$.
    Solution

  • Question 6
    1 / -0
    Which of the following processes will increase $$[OH^{-}]$$ in $$NH_{4}OH$$ solution?
    Solution

  • Question 7
    1 / -0
    Which of the following statement(s) is/are incorrect?
    Solution

  • Question 8
    1 / -0
    The amino acid alanine has two isomers, $$ \alpha $$ alanine and $$ \beta $$ alanine when equal masses of these two compound are dissolved in equal mass of a solvent, the solution of $$ \alpha $$ alanine freeze at relatively l; lower temperature, which forms $$ \alpha $$ alanine and $$ \beta $$ alanine has the larger equilibrium constant for ionization?
    Solution
    $$\alpha\space alanine$$ and $$\beta\space alanine$$ both dissociate into two ion as shown below,

    $$Alanine\leftrightharpoons X^++Y^-$$

    Equilibrium constant is ratio of the concentration of product and concentration of reactant.
    So, $$K_{c(alanine)}=[X^+][Y^-]$$

    The freezing point of the solvent decreases when a non-volatile solute is added to pure solvent this is known as depression in freezing point.
    Ions in solution change the freezing point temperature.

    If more ions are present in the solution, they decrease the freezing point more.
    Since it's given that $$\alpha\space alanine$$ has a lower freezing temperature it implies that $$\alpha\space alanine$$  has more ions.
    Therefore, the equilibrium constant of $$\alpha\space alanine$$ is larger.

    Hence, the correct option is (A).

  • Question 9
    1 / -0
    The initial rate of hydrolysis of methyl acetate $$(1.0M)$$ by a weak acid $$(HA,1.0M)$$ is 1/100th of that of a strong acid $$(HX,1.0M)$$, at $${ 25 }^{ o }C$$. The $${K}_{a}$$ of $$HA$$ is
    Solution

  • Question 10
    1 / -0
    Solid ammonium carbamate dissociates as: $$ NH_2COONH_{4_{(s)}} \rightleftharpoons  2NH_{3_{(g)}} + CO_{2_{ (g)}} $$ 

    In a closed vessel, solid ammonium carbamate is in equilibrium with its dissociation products. At equilibrium, ammonia is added such that the partial pressure of $$ NH_3 $$ at new equilibrium equals the original total pressure. The ratio of total pressure at new equilibrium to that of original total pressure is:
    Solution

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