How Do You Calculate the Solubility of AgBrO3 from Given Data?

[lo2]

I have this salt $$AgBrO_{3}$$ which has got the solubility of $$\frac{0,196g}{100mL}$$. From the following information I should be able to calculate Solubility for $$AgBrO_{3}$$.

I really cannot see how? So please help me.

It is really an Urgent matter so a quick replay would be most appreciated!

 

[dmoravec]

unless I am completely misreading it I think you already have your answer. Maybe you are expecting different units? [g/L, mol/L, etc]

 

[lo2]

unless I am completely misreading it I think you already have your answer. Maybe you are expecting different units? [g/L, mol/L, etc]

No I want to find the solubility product constant...

 

[dmoravec]

well the solubilty constant is much like any other equilibrium constant.

A reaction of the form

AB{s} -> xA(+){aq} + yB(-){aq} where the x and y are balanced coefficients, and the species A(+) and B(-) are the charged ions ( not necessarily of charge 1)

Then the solubility constant ( call it K(s)) is

K(s) = ([A(+)]^x) * ([B(-)]^y)

where [A(+)] and [B(-)] are the concentrations of species A and B. Do you have the reaction formula for the dissociation of Silver Bromate?

 

[lo2]

well the solubilty constant is much like any other equilibrium constant.

A reaction of the form

AB{s} -> xA(+){aq} + yB(-){aq} where the x and y are balanced coefficients, and the species A(+) and B(-) are the charged ions ( not necessarily of charge 1)

Then the solubility constant ( call it K(s)) is

K(s) = ([A(+)]^x) * ([B(-)]^y)

where [A(+)] and [B(-)] are the concentrations of species A and B. Do you have the reaction formula for the dissociation of Silver Bromate?

I know that. That is not the problem.

My problem is how to find the Solubilty product constant, from that one and only piece of information I have got, which is that the solubility for $$AgBrO_{3}$$ is $$\frac{0,196g}{100mL}$$. That is the problem!

 

[dmoravec]

remember, you can use the molecular weight of silver bromate (235.77 g/mol) to figure out the amount of moles/L of both the silver and bromate ions that were dissolved. From there these values can be plugged in as A(+) and B(-) with x and y both being 1.

 

[lo2]

I do not quite think that I understand.

 

[dmoravec]

ok
Silver Bromate dissocates into and silver ion and bromate ion as follows:

AgBrO(3) -> Ag(+) + BrO3(-).

By the formula for solubility constants the constant for this reaction would be

K(s) = [Ag(+)]*[BrO3(-)].
So all we need to get is the concentration (in moles/L) of these ions. You have the information of g/mL. Specifically 0.196 g/mL. To convert that to an equivalent moles/L first convert the amount of mass dissolved (0.196g) into moles.

0.196 g / 235.77 g/mol = 0.000831 moles.
Now this is in 100 mL or 0.1 L.
Therefore the amount of moles dissolved / liter is
0.000831 moles / 0.1 L = 0.00831 moles/L (commonly referred to as M)

From the above equation we know that for each mole of silver bromate dissolved there is exactly 1 mole of silver AND 1 mole of bromate. Therefore the concentration of silver ([Ag(+)]) is 0.00831 moles/L and likewise the concentration of bromate ions is the same.

Therefore the solubility constant is:
0.00831 * 0.00831 = Answer

a different, but similar, example can be seen here:
http://faculty.kutztown.edu/vitz/limsport/LabManual/KSPWeb/KSP.htm

 

[lo2]

Wow some explanation!

I owe you a favour just request and I will help you out if I can.