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Suppose $\ce{Na2SO4}$ has a conductivity $\pu{2.6 \times 10^{-4} S cm-1}$ and on mixing $\ce{CaSO4}$ to saturation, the conductivity becomes $\pu{7.0 \times 10^{-4} S cm-1}$.

So, Is the conductivity of $\ce{CaSO4}$ the difference of $\pu{7.0 \times 10^{-4} S cm-1}$ and $\pu{2.60 \times 10^{-4} S cm-1}$?

I got an answer of a similar question here. But It is explaining things opposite to what I said to get the desired answer.

My book also solved by subtracting the two conductivities.

So I am confused, Please help me reach out an answer.

Subsrtaction of conductivities

This question is a part of a large question.

This is written in my book.


EDIT:- I am adding the original question for further clarity

Original question

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    $\begingroup$ You can edit this question, and it is needed. Please take a site tour. $\endgroup$
    – Todd Minehardt
    Jul 26 at 17:58
  • $\begingroup$ At this level of dilution (calcium sulfate is only sparingly soluble, so it has little impact unless the initial sodium sulfate is also very dilute), you can just take a difference. More concentrated solutions you would get with salts having good solubility are complicated by ion-ion interactions. $\endgroup$ Jul 26 at 17:59
  • $\begingroup$ As for editing, I would describe the large question to provide context. Such context will allow us to develop a good answer. $\endgroup$ Jul 26 at 18:05
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    $\begingroup$ @Oscar Lanzi I've added the original question $\endgroup$ Jul 26 at 18:21
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    $\begingroup$ You are missing one critical point. All these additions and subtractions of molar conductivities are valid at infinite dilution. This is called Kohlrausch's law. If you know the molar conductivity of NaCl at infinite dilution and then you have a saturated $\ce{CaSO4}$ and it is still such a dilute solution that you can consider it at "infinite dilution". Your excercise book is using that approximation during subtraction. $\endgroup$
    – M. Farooq
    Jul 26 at 21:46
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OP's question is: Is the conductivity of $\ce{CaSO4}$ the difference of $\pu{7.0 \times 10^{−4} S cm−1}$ and $\pu{2.6 \times 10^{−4} S cm−1}$?

In this question, $\pu{2.6 \times 10^{−4} S cm−1}$ is the conductivity of original aqueous $\ce{Na2SO4}$ solution and $\pu{7.0 \times 10^{−4} S cm−1}$ is the conductivity of original aqueous solution when saturated with solid $\ce{CaSO4}$. The initial question before edition, it doesn't make sense because of the lack of information (e.g., no concentration terms were given). I completely agree with Oscar Lanzi and M. Farooq with their valuable comments.

Now, since OP has edited the question and given the full description of main question, we can give an approach to solve the question. It is evident that this is not a high level of analytical question (which would attract M. Farooq). Therefore we can take certain assumption such as the solutions are dilute enough to be considered as those of infinitely dilute solutions.

Strong electrolytes are hypothesized to dissociate completely in solution. The conductivity of a solution of a strong electrolyte at low concentration follows Kohlrausch's Law (see M. Farooq's comment: You are missing one critical point. All these additions and subtractions of molar conductivities are valid at infinite dilution. This is called Kohlrausch's law. If you know the molar conductivity of $\ce{NaCl}$ at infinite dilution and then you have a saturated $\ce{CaSO4}$ and it is still such a dilute solution that you can consider it at "infinite dilution". Your exercise book is using that approximation during subtraction):

$$\Lambda_m = \Lambda^\circ_m - K \sqrt{c}$$ where $\Lambda_m$ is molar conductivity, $\Lambda^\circ_m$ is limiting molar conductivity (at the limit of the infinite dilution), $K$ is an empirical constant, and $c$ is the electrolyte concentration. At sufficiently low concentrations solutions (when $\Lambda^\circ_m \gt\gt K \sqrt{c}$), the observed conductivity of a strong electrolyte becomes directly proportional to concentration of the solution. As quoted by Wikipedia:

The specific conductance of a solution containing one electrolyte depends on the concentration of the electrolyte. Therefore, it is convenient to divide the specific conductance by concentration. This quotient, termed molar conductivity, is denoted by $\Lambda_m$, $$\Lambda_m = \frac{\kappa}{c} \tag1$$

If the limiting molar conductivities of cation and anion are $\lambda^\circ_+$ and $\lambda^\circ_-$, for the diluted enough solutions, $\lambda_+ \approx \lambda^\circ_+$ and $\lambda_- \approx \lambda^\circ_-$, and $$\kappa = c_+\lambda_+ + c_-\lambda_- \tag2$$

Now, let's look at the question in hand. Since $\pu{2.6 \times 10^{−4} S cm−1}$ is the conductivity of original aqueous $\pu{0.001 mol L-1}$ $\ce{Na2SO4}$ solution (or $\pu{0.001 \times 10^{-3} mol cm-3}$): $$\ce{Na2SO4 -> 2Na+ + SO4^2-}$$ and from the equation $(2)$: $$\pu{2.6 \times 10^{−4} S cm−1} = c_\ce{Na+}\lambda_\ce{Na+} + c_\ce{SO4^2+}\lambda_\ce{SO4^2+} \tag3$$

$$\pu{2.6 \times 10^{−4} S cm−1} = 2 \times \pu{0.001 \times 10^{-3} mol cm-3} \times \pu{50 S cm2 mol-1} + \pu{0.001 \times 10^{-3} mol cm-3} \times \lambda_\ce{SO4^2-} $$ $$\therefore \ \lambda_\ce{SO4^2-} = \pu{160 S cm2 mol-1}$$

Similarly, when the solution is saturated with $\ce{CaSO4}$: $$\ce{CaSO4 <=> Ca^2+ + SO4^2-}$$ And, from the equation $(2)$ for the diluted mixed solution (suppose $\alpha$ amount of $\ce{CaSO4}$ has dissolved): $$\kappa = c_\ce{Na+}\lambda_\ce{Na+} + c_\ce{SO4^2+}\lambda_\ce{SO4^2+} + c_\ce{Ca^2+}\lambda_\ce{Ca^2+} + c_\ce{SO4^2+}\lambda_\ce{SO4^2+}\tag4$$

Applying the equation $(3)$ and $\lambda_\ce{SO4^2-} = \pu{160 S cm2 mol-1}$ in the equation $(4)$ (avoid the units for convenience):

$$7.0 \times 10^{−4} = 2.6 \times 10^{−4} + \alpha \times 120 + \alpha \times 160 \tag4$$ $$\therefore \ \alpha = \frac{7.0 \times 10^{−4} - 2.6 \times 10^{−4}}{120+160} = \pu{1.57 \times 10^{−6} mol cm-3} = \pu{1.57 \times 10^{−3} mol L-1}$$

If you take $K_\mathrm{sp} = \alpha^2 = 2.46 \times 10^{−6}$, there is an answer, but it was incorrect (that is the common trick for multiple choice questions). That's because there is a common ion effect you may need to consider here. Thus correct answer for $K_\mathrm{sp}$ is:

$$K_\mathrm{sp} = \alpha(\alpha + 0.001) = 2.46 \times 10^{−6} + 1.57 \times 10^{−3} \times 0.001 = 4.03 \times 10^{−6} \approx 4.0 \times 10^{−6}$$

Thus, correct answer is choice $(a)$.

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