Question

Consider the reaction
CaSO4(s)⇌Ca2+(aq)+SO2−4(aq) 
At 25 ∘C the equilibrium constant is Kc=2.4×10−5 for this reaction.
- If excess CaSO4(s) is mixed with water at 25 ∘C to produce a saturated solution of CaSO4, what is the equilibrium concentration of SO2−4?
- If the resulting solution has a volume of 1.2 L, what is the minimum mass of CaSO4(s) needed to achieve equilibrium?

Answer 1

a) According to the reaction equation:

CaSO4(s) ↔ Ca2+(aq)  + SO4-(aq)

when Kc = [Ca2+][SO4-]

when [Ca2+] = [SO4-] = X

and we have Kc = 2.4 x 10^-5 so ,by substitution we can get [Ca2+]&[SO4-] 

2.4 x 10^-5 = X^2

∴X = 0.0049

∴[SO4-] = [Ca2+] = 0.0049 M

b) when we have [Ca2+] =  0.0049 M so we can get the no.of moles of CaSO4:

moles of CaSO4 = molarity * volume 

                             = 0.0049 M * 1.2 L
                             = 0.00588 moles

when we know the molar mass of CaSO4 = 136.14 g/ mol, So we can get the mass:

∴mass of CaSO4 = moles of CaSO4 * molar mass of CaSO4

                             = 0.00588 moles * 136.14 g/mol

                             = 0.8 g

Answer 2

The equilibrium concentration of ${\text{SO}}_4^{2 - }$ is $\boxed{{\text{0}}{\text{.004899 M}}}$.

The minimum mass of ${\text{CaS}}{{\text{O}}_{\text{4}}}$ required to attain the given equilibrium is $\boxed{0.8{\text{ g}}}$.

Further Explanation:

Chemical equilibrium refers to a stage where the rate of both forward and backward reactions is same.

The generic equation for a general equilibrium is as follows:

$a{\text{A}} + b{\text{B}} \rightleftharpoons c{\text{C}} + d{\text{D}}$  

Where,

A and B are the reactants.

C and D are the products.

a and b are the stoichiometric coefficients of reactants.

c and d are the stoichiometric coefficients of products.

The formula to evaluate the equilibrium constant for the above reaction is,

${K_{\text{c}}} = \dfrac{{{{\left[ {\text{C}} \right]}^c}{{\left[ {\text{D}} \right]}^d}}}{{{{\left[ {\text{A}} \right]}^a}{{\left[ {\text{B}} \right]}^b}}}$  

Here,

${{\text{K}}_{\text{c}}}$ is the equilibrium constant.

[C] is the concentration of C.

[D] is the concentration of D.

[A] is the concentration of A.

[B] is the concentration of B.

The given reaction is as follows:

${\text{CaS}}{{\text{O}}_{\text{4}}}\left( s \right) \rightleftharpoons {\text{C}}{{\text{a}}^{2 + }}\left( {aq} \right) + {\text{SO}}_4^{2 - }\left( {aq} \right)$

The concentration of pure solid is zero. Therefore the expression for equilibrium constant for the above equation is as follows:

${{\text{K}}_{\text{c}}} = \left[ {{\text{C}}{{\text{a}}^{2 + }}} \right]\left[ {{\text{SO}}_4^{2 - }} \right]$   …… (1)                                                                            

Here,

$\left[ {{\text{C}}{{\text{a}}^{2 + }}} \right]$ is the concentration of ${\text{C}}{{\text{a}}^{2 + }}$ ions.

$\left[ {{\text{SO}}_4^{2 - }} \right]$ is the concentration of ${\text{SO}}_4^{2 - }$ ions.

Let us assume the concentration of both ${\text{C}}{{\text{a}}^{2 + }}$ and ${\text{SO}}_4^{2 - }$ ions to be x. Substitute x for $\left[ {{\text{C}}{{\text{a}}^{2 + }}} \right]$ , x for $\left[ {{\text{SO}}_4^{2 - }} \right]$ and $2.4 \times {10^{ - 5}}$ for ${{\text{K}}_{\text{c}}}$ in equation (1).

$\begin{aligned}2.4 \times {10^{ - 5}} &= \left( x \right)\left( x \right) \hfill\\{x^2} &= 2.4 \times {10^{ - 5}} \hfill\\x &= \sqrt {2.4 \times {{10}^{ - 5}}}  \hfill\\\end{aligned}$  

Solving for x,

$x = 0.004899$

Therefore the equilibrium concentration of ${\text{SO}}_4^{2 - }$ is 0.004899 M.

Since one mole of ${\text{C}}{{\text{a}}^{2 + }}$ is formed by the dissociation of one mole of ${\text{CaS}}{{\text{O}}_{\text{4}}}$. Therefore the concentration of ${\text{CaS}}{{\text{O}}_{\text{4}}}$ is 0.004899 M.

The formula to calculate the molarity of ${\text{CaS}}{{\text{O}}_{\text{4}}}$ solution is as follows:

${\text{Molarity of CaS}}{{\text{O}}_{\text{4}}}{\text{ solution}} = \dfrac{{{\text{Amount of CaS}}{{\text{O}}_{\text{4}}}}}{{{\text{Volume of CaS}}{{\text{O}}_{\text{4}}}}}$          ……. (2)                                          

Rearrange equation (2) for the amount of ${\text{CaS}}{{\text{O}}_{\text{4}}}$.

${\text{Amount of CaS}}{{\text{O}}_{\text{4}}} = \left( {{\text{Molarity of CaS}}{{\text{O}}_{\text{4}}}{\text{ solution}}} \right)\left( {{\text{Volume of CaS}}{{\text{O}}_{\text{4}}}} \right)$                   …… (3)

Substitute 0.004899 M for the molarity and 1.2 L for the volume of   in equation (3).

 $\begin{aligned}{\text{Amount of CaS}}{{\text{O}}_{\text{4}}} &= \left( {{\text{0}}{\text{.004899 M}}} \right)\left( {{\text{1}}{\text{.2 L}}} \right)\\&= 0.0058788{\text{ mol}}\\\end{aligned}$

The formula to calculate the mass of ${\text{CaS}}{{\text{O}}_{\text{4}}}$ is as follows:

${\text{Mass of CaS}}{{\text{O}}_{\text{4}}} = \left( {{\text{Moles of CaS}}{{\text{O}}_{\text{4}}}} \right)\left( {{\text{Molar mass of CaS}}{{\text{O}}_{\text{4}}}} \right)$                   …… (4)                  

Substitute 0.0058788 mol for the moles and 136.14 g/mol for the molar mass of ${\text{CaS}}{{\text{O}}_{\text{4}}}$ in equation (4).

 $\begin{aligned}{\text{Mass of CaS}}{{\text{O}}_{\text{4}}} &= \left( {{\text{0}}{\text{.0058788 mol}}} \right)\left( {\frac{{{\text{136}}{\text{.14 g}}}}{{1{\text{ mol}}}}} \right)\\&=0.800339832\;{\text{g}}\\&\approx {\text{0}}{\text{.8}}\;{\text{g}}\\\end{aligned}$

Learn more:

1. Calculate equilibrium constant for ammonia synthesis: https://brainly.com/question/8983893
2. Complete equation for the dissociation of  (aq): https://brainly.com/question/5425813

Answer details:

Grade: Senior School

Chapter: Chemical Equilibrium

Subject: Chemistry

Keywords: chemical equilibrium, Kc, CaSO4, Ca2+, SO42-, 0.8 g, 136.14 g/mol, 0.004899 M.