(a)
Interpretation:
The balanced dissociation reaction equation and the associated equilibrium constant expression for the given processes has to be written.
Concept Introduction:
Balanced Chemical equation:
A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.
Equilibrium constant: At equilibrium the ratio of products to reactants (each raised to the power corresponding to its
For a general reaction,
The concentration of solids and pure liquids do not change, so their concentration terms are not included in the equilibrium constant expression.
(b)
Interpretation:
Among the given compounds the one that is more soluble has to be given along with explanation.
(c)
Interpretation:
Among the given compounds the one that is less soluble has to be given along with explanation.
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Chapter 12 Solutions
Chemistry: The Molecular Science
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- 3. Sulphuric acid (H2SO4 ), used in the manufacture of fertilizers, undergo decomposition and the acid dissociation constants are given as: Ka1 = 1x 10° and K.2 = 1.2 x 10¬2 . The initial concentration of sulphuric acid is given as 0.040M. (i) Write the dissociation reactions. (ii) Calculate the concentration of [HS0, ] and [SO]. (iii) Calculate the pH of the acid solution. (iv) Write a discussion on the behaviour of polyprotic acids. (50 -100 words) Given: lon product constant for water, K, = 1 x 10-14arrow_forward3. When combined in a closed vessel, hydrogen gas and iodine gas will form hydrogen iodide gas until an equilibrium position is reached. m MO (a) Write a balanced equation for this chemical reaction system. (b) Suppose you carry out an investigation starting with a 2.0 L flask containing 0.45 mol of hydrogen iodide. Predict how the concentrations of the gases will change as the system reaches equilibrium. (c) In a second experiment, the concentration of each of the gases was monitored (Figure 9). Complete an ICE table for the reaction. Reaction Progress 8.0 t= 200 °C 7.0 (H(g) 6.0 5.0 4.0 3.0 - [H(9)] 2.0 1.0 0.0 Time Figure 9 Concentration (mol/L)arrow_forwardConsider the equilibrium system described by the chemical reaction below. If 8.98 × 104 moles of KOH(aq), 8.98 × 104 moles of HCN(aq), and 0.0500 moles of KCN(aq) are present in a 1.00 L aqueous solution at equilibrium at 298 K, what is the value of K of the reaction at this temperature? KCN(aq) + H,O(I) = KOH(aq) + HCN(aq) [8.98 × 104] K || [0.0500] [1.80 x 10-²] 1 [8.06 x 10-7] 1.80 x 10-² RESET 1.61 x 10-5arrow_forward
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