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Equilibrium Test - 13

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Equilibrium Test - 13
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  • Question 1
    1 / -0
    Which of the following mixture is not a buffer solution?
    Solution
    Basic buffer solution is prepared by mixing a weak base and its salt with strong acid.
    In the mixture containing 100ml of 0.5 N $$NH_4OH$$ + 100ml of 0.5 N-HCl, equal number of moles of ammonium hydroxide and hydrochloric acid are present.
    They completely neutralize each other to form salt ammonium chloride. Thus weak base ammonium hydroxide is not present in the mixture. 
    Hence, this mixture is not a buffer solution.
  • Question 2
    1 / -0
    1M NaCl and 1M HCl are present in an aqueous solution . The solution is:
    Solution
    Buffer can accept and donate protons at the same time and $$HCl$$ is an acid. So, it has $$pH<7$$.
    So, this is not a buffer and the solution will be acidic.

    Hence, the correct option is $$A$$
  • Question 3
    1 / -0
    The $$pH$$ of a solution obtained by mixing $$50\: ml$$ of $$0.4N$$ $$NH_{4}Cl$$ and $$50\: ml$$ of $$0.2 N$$ $$NaOH$$ is:
    Solution
    The expression for the hydrogen ion concentration is as given below.

    $$[H^+]= \dfrac {V_bN_b-V_aN_a} {V_a+V_b}$$

    Substitute values in the above expression,

    $$[H^+]= \dfrac {50 \times 0.4-50 \times0.2} {50+50}=0.1$$

    $$pH=-log\ [H^+]=-log\ (0.1)=1$$
  • Question 4
    1 / -0
    Assertion: A buffer solution has a capacity to resist the change in pH value on an addition of a small amount of acid or base to it.
    Reason: An aqueous solution of ammonium acetate can act as buffer.
    Solution
    salts of weak-acid and weak-base can act as buffer solution.
  • Question 5
    1 / -0
    Assertion: A mixture of $$CH_3COOH$$ and $$CH_3COONH_4$$ is not a buffer solution.
    Reason: An acidic buffer contains the mixture of a weak acid and its salt with a strong base.
    Solution
    Acidic buffer solution is prepared by mixing a weak acid and its salt with strong base.  A solution containing weak acid and its salt with weak base is not a buffer solution.
    Ammonium acetate is a salt of weak acid and weak base. Hence, mixture of acetic acid and ammonium acetate is not a buffer solution.
  • Question 6
    1 / -0
    $$C_{2}H_{4}(g) + H_{2}(g)\leftrightharpoons  C_{2}H_{6}(g);\ \Delta H = - 31.7 \ kcal$$.
    The equilibrium concentration of $$C_{2}H_{4}$$ in the following reaction can be increased by:
    Solution
    Negative value of the enthalpy change $$(\Delta H = - 31.7 K)$$ for the reaction $$C_{2}H_{4}(g) + H_{2}(g)\leftrightharpoons  C_{2}H_{6}(g)$$ indicates that it is an exothermic reaction.

    When the temperature of an exothermic reaction is increased, the reaction will proceed in the backward direction so as to absorb heat and nullify the effect of increasing temperature. This will decrease the value of the equilibrium constant $$K_c$$. In other words, the equilibrium will shift to left.

    Thus, to increase the equilibrium concentration of $$C_{2}H_{4}$$, the temperature should be increased.
  • Question 7
    1 / -0
    The equilibrium constant of a reaction at $$298 \:K$$ and $$1000\: K$$ is $$5 \times 10^{-3}$$ and $$2\times 10^{-3}$$ respectively. The $$\Delta  H$$ for the reaction is:
    Solution
    With the increase in temperature from $$298 \:K$$ to $$1000 \:K$$, the value of the equilibrium constant decreases from $$5 \times 10^{-3}$$ to $$2\times 10^{-3}$$.

    Thus, as the temperature increases, the equilibrium shifts to the left direction. This is possible for an exothermic reaction.

    Thus, the reaction is an exothermic reaction having a negative value of enthalpy change.
  • Question 8
    1 / -0
    The reaction $$A + B \leftrightharpoons C + D + heat$$, has reached equilibrium. It is possible to make the reaction to proceed forward:
    Solution
    The reaction $$A + B \Leftrightarrow C + D + heat$$ is an exothermic reaction. 

    According to Le Chatelier's principle, as the temperature is lowered, the reaction will proceed in the direction so as to liberate heat and nullify the effect of lowering the temperature.

    Hence, the reaction will proceed in the forward direction by lowering the temperature.
  • Question 9
    1 / -0
    Assertion: pH of blood does not change in spite of taking acidic foods.
    Reason: Blood behaves as a buffer solution.
    Solution
    Blood contains a mixture of $$H_{2}CO_{3}$$ and $$HCO_{3}^{-} $$ which acts as acidic buffer solution.
    Thus blood has the tendency to resist changes in pH when small quantities of acid or base are added.
  • Question 10
    1 / -0
    In the system $$CaF_{2} (s) \rightleftharpoons Ca^{2+}(aq) + 2F^{-}(aq)$$, if the concentration of $$Ca^{2+}$$ is increased by four times, then the equilibrium concentration of $$F^{-}$$ will be changed to:
    Solution
    The equilibrium reaction is as follows:
    $$CaF_{2} (s) \rightleftharpoons Ca^{2+}(aq) + 2F^{-}(aq)$$.

    The expression for the equilibrium constant is $$K=[Ca^{2+}][F^-]^2$$.

    Let $$X$$ and $$Y$$ be the initial concentrations of $$[Ca^{2+}]$$ and $$[F^-]$$ ions.

    The expression for the equilibrium constant will be $$K=[Ca^{2+}][F^-]^2=XY^2$$ ......$$(1)$$

    When the concentration of the $$Ca^{2+}$$ ions is increased four times, the expression for the new equilibrium constant will be $$K=[Ca^{2+}][F^-]^2=4X\times [F^-]^2$$ ...... $$(2)$$

    The magnitude of the equilibrium constant is not affected by the changes in concentrations of reactants and products.

    Thus, from $$(1)$$ and $$(2)$$, we get $$K=XY^2=4X\times [F^-]^2$$
    $$\implies [F^-]= \dfrac {Y} {2}$$.

    Thus, the equilibrium concentration of $$F^{-}$$ ions is reduced to one half its original value.

    Hence, the correct option is $$\text{A}$$
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