Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Need some help on some Chemistry calculations

  1. Aug 23, 2006 #1


    User Avatar

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

    I really cannot see how? So please help me.

    It is really an Urgent matter so a quick replay would be most appreciated!
  2. jcsd
  3. Aug 23, 2006 #2
    unless I am completely misreading it I think you already have your answer. Maybe you are expecting different units? [g/L, mol/L, etc]
  4. Aug 23, 2006 #3


    User Avatar

    No I want to find the solubility product constant...
  5. Aug 23, 2006 #4
    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?
  6. Aug 23, 2006 #5


    User Avatar

    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 [tex]AgBrO_{3}[/tex] is [tex]\frac{0,196g}{100mL}[/tex]. That is the problem!
  7. Aug 23, 2006 #6
    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.
  8. Aug 23, 2006 #7


    User Avatar

    I do not quite think that I understand.
  9. Aug 23, 2006 #8
    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 refered 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 [Broken]
    Last edited by a moderator: May 2, 2017
  10. Aug 23, 2006 #9


    User Avatar

    Wow some explanation!

    I owe you a favour just request and I will help you out if I can.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook