Video Transcript
Thallium salts are highly toxic with very low solubilities. However, it was possible to dissolve 3.49 grams of thallium(I) bromate, TlBrO3, in
one liter of water during a laboratory preparation. What is the solubility product of thallium(I) bromate? Taking the molar mass of thallium(I) bromate to be 332.3 grams per mole, give your
answer in scientific notation to three decimal places.
We generally think of solids as being soluble or insoluble. But many solids we consider insoluble simply have very low solubilities, meaning that
they do dissolve but only a little bit. We can express this dissolution as an equilibrium reaction. Here is a general reaction for the dissolution of a general metal salt.
We can define an equilibrium constant for this reaction, which is called the
solubility product. It is equal to the concentration of the metal cation times the concentration of the
anion. The concentration indicated by square brackets is in units of molar, moles per liter,
or moles per cubic decimeter. If there are stoichiometric coefficients involved in the reaction, we can include
that in the solubility product by raising each concentration to the power of the
stoichiometric coefficient.
Let’s define the solubility product for the metal salt in this question, thallium(I)
bromate. When thallium(I) bromate dissolves, it dissociates into thallium one plus ions and
bromate anions. So the 𝐾 sp for thallium bromate is equal to the concentration of thallium ions
times the concentration of bromate ions. So we’re going to need the concentration of thallium and bromate ions to solve this
problem.
We don’t know the concentrations. That information wasn’t given in the problem. But we do know that every mole of thallium bromate dissociates to form one mole of
thallium ions and one mole of bromate ions. For every amount X of thallium bromate that dissolves in the solution, that same
amount X of both thallium and bromate ions will be formed. So if we calculate the amount of thallium bromate that dissolves in the solution, we
can calculate the solubility product.
The problem tells us that 3.49 grams of thallium(I) bromate dissolves into the
solution. But the amount we want needs to be expressed in moles, so we’ll have to convert from
grams to moles. We can find the amount of a substance in moles by dividing the mass of the substance
by its molar mass. So let’s go ahead and calculate the amount of thallium bromate in moles.
The mass of thallium bromate given in the problem is 3.49 grams. And the molar mass was given to be 332.3 grams per mole. Performing the calculation gives us the amount of thallium(I) bromate that
dissolved. So the amount of thallium(I) bromate is 0.0105 moles. Because of the stoichiometry of the reaction, this amount is also the amount of
thallium one plus ions and bromate ions.
Now all we need to do before we solve the problem is turn this amount into a
concentration. We can calculate the concentration if we divide the amount of a substance in moles by
the volume of the solution. The volume of the solution is one liter. So the concentration of thallium one plus and bromate ions are both equal to 0.0105
moles per liter.
So finally, we have what we need to solve for the 𝐾 sp. The concentrations of thallium one plus ions and bromate ions are both equal to
0.0105 moles per liter, which is equal to the concentration squared. Performing the calculation gives us 1.1025 times 10 to the power of negative four
moles squared per liter squared. If we round to three decimal places, we’ll have our final answer. So the solubility product of thallium(I) bromate is 1.103 times 10 to the power of
negative four moles squared per liter squared.
