Order the following ionic compounds in decreasing order according to their ease to dissolve in water.

a) NaBr; b) KI; c) NaF; d) NaCl.

  1. d > a > b > c
  2. a > b > c > d
  3. b > c > d > a
  4. c > d > a > b

So, according to what I researched the solubility $s$ for each one of these compounds in water (at $\pu{20 °C})$ is:

$$ \begin{array}{lc} \hline \text{Compound} & s/\pu{g L-1}\\ \hline \ce{NaBr} & 905 \\ \ce{KI} & 1400 \\ \ce{NaF} & 40.4 \\ \ce{NaCl} & 359 \\ \hline \end{array} $$

Therefore, my answer would be, from more soluble to least soluble:

b) KI > a) NaBr > d) NaCl > c) NaF

But this answer is not among the list. I think option 4 might be "correct", but it's kind of backwards.

Are the given answers wrong or am I doing something wrong?

  • 1
    $\begingroup$ You know you're supposed to be able to answer this question without looking up the numbers, right? And it doesn't matter if it's molar or mass solubility. You only need to know one little fact. $\endgroup$
    – Karl
    Dec 6 '20 at 20:02
  • $\begingroup$ Teach me please, or tell me where to go. I'd like to know because these online classes are sooo bad. $\endgroup$
    – Adolf
    Dec 7 '20 at 0:24
  • $\begingroup$ NaF is not very well soluble. The halogens Cl, Br, I are rather similar in behaviour, flourine and the flourides (nearly) always stick out. You cannot be expected to know those numbers by heart, you know that NaCl is very soluble, so the outlier must be NaF, on the lower end. $\endgroup$
    – Karl
    Dec 7 '20 at 9:50

The order b-a-d-c is correct, if the solubility is in gramm per liter. If it is in mole per liter, the order is a>b>c>d, as the saturated solutions of NaBr is the highest ($8.78~ mol L^{-1}$), then KI ($8.48~ mol L^{-1}$), then NaCl ($6.19~ mol L^{-1}$), and NaF ($0.95~ mol L^{-1}$)


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