Question

Based on the average number of moles of IO3- in your samples, and the initial volume of those samples, what is the solubility of Ca(IO3)2 in mol/L of the saturated solution in 0.01 M KIO3?

Answer 1

Answer : The solubility of $Ca(IO_3)_2$ is, 0.0036 M

Explanation :

First we have to calculate the concentration of $IO_3^-$

$\text{Concentration of }IO_3^-=\frac{\text{Moles of }IO_3^-}{\text{Volume of solution}}$

$\text{Concentration of }IO_3^-=\frac{0.000086mol}{5mL}=\frac{0.000086mol\times 1000}{5L}=0.0172M$

Now we will subtract $IO_3^-$  that came from $KIO_3$

Concentration of $IO_3^-$ = 0.0172 - 0.01 = 0.0072 M

Now we have to calculate the solubility of $Ca(IO_3)_2$

$Ca(IO_3)_2\rightarrow Ca^{2+}+2IO_3^-$

                     s           2s

$[IO_3^-]$ = 0.0072 M

2s = 0.0072 M

s = 0.0036 M

Thus, the solubility of $Ca(IO_3)_2$ is, 0.0036 M