## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 15 - Exercises - Page 715: 49

#### Answer

$Q_c = K_c$. Therefore, the reaction mixture is exactly at equilibrium, and the solution is saturated. If we add more solid silver sulfate it will $not$ dissolve.

#### Work Step by Step

1. Determine the initial concentration of the ions. $Ag_2SO_4$ : ( 107.9 $\times$ 2 )+ ( 16.00 $\times$ 4 )+ ( 32.07 $\times$ 1 )= 311.9 g/mol - Calculate the amount of moles: $$6.55 \space g \times \frac{1 \space mol}{ 311.9 \space g} = 0.0210 \space mol$$ - Calculate the molarity: $$\frac{ 0.0210 \space mol}{ 1.5 \space L} = 0.014 \space M$$ If all the substance dissolved: $[Ag^{+}] = 2*0.014 \space M = 0.028 \space M$ $[S{O_4}^{2-}] = 0.014 \space M$ 2. Determine the $Q_c$ of the reaction mixture: - The exponent of each concentration is equal to its balance coefficient. $$Q_C = \frac{[Products]}{[Reactants]} = [ Ag^{+} ] ^{ 2 }[ S{O_4}^{2-} ]$$ 3. Substitute the values and calculate the constant value: $$Q_C = ( 0.028 )^{ 2 }( 0.014 ) = 1.1 \times 10^{-5}$$ $Q_c = K_c$. Therefore, the reaction mixture is exactly at equilibrium, and the solution is saturated. If we add more solid silver sulfate it will not dissolve.

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