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: Long Titration Problem

  1. Nov 10, 2007 #1
    This was a pretty long double titration problem, I solved it all except for the last part & could use a bit of help because I'm stuck.

    Part 1 A 100.00 ml solution containing aqueous HCl & HBr was titrated with .1200 M NaOH. The volume of base required to neutralize the acid was 47.14 mL. a) Write the net ionic equation for both reactions that occur during the titration.
    Answer: 2H+ (aq) + 2(OH)- (aq) -> 2H2O (l)

    Part 2 Determine the total number of moles of acid (HBr + HCl) in the original solution.
    Answer: .002828 mol HBr & .002828 mol HCl

    Part 3 Aqueous AgNO3 was then added to the solution (after the titration). Write the net ionic equation for both reactions that occur.
    Answer: Br- (aq) + Cl- (aq) + 2Ag+ (aq) -> AgBr (s) + AgCl (s)

    Part 4 The total mass of solid produced from the above reactions was .9975 g. Did the original solution contain more HCl or more HBr (in grams)?
    This is the part I'm stuck on. At first I tried to take the mass percents of both of the solid products, into the total mass of the solid produced. But when I talked to my professor, she stated there wasn't necessarily equal moles of HCl & HBr in the beginning. Any ideas?
  2. jcsd
  3. Nov 11, 2007 #2
    algebra, 2 equations, 2 unknowns

    your professor is right that there is nothing in the problem that says there are equal numbers of moles of HCl and HBr

    so the most you can do is say x mol HCl + y mol HBr = .0056568 moles

    1. x mol Cl- + y mol Br- = .0056568 mol

    that is your first equation with 2 unknowns

    the second comes from the ppt with AgNO3

    the two products AgBr and AgCl TOGETHER give a combined mass of .9975g

    so g AgBr + g AgCl = .9975 g

    but this doesn't have the same x and y as the first equation, so change g into something that has mol

    MM = g/mol so instead of g make it

    MM AgBr * y mol AgBr (write out the net ionic equation for AgBr)

    or this is the same as g AgBr = MM AgBR * y mol Br-

    (write out the net ionic equation for AgBr)

    do the same for AgCl, you get g AgCl = MM AgCl * x mol Cl-

    put this back in to the grams equation in green and this becomes equation #2

    then its an algebra problem... good luck

    I probably did too much for you...
    Last edited: Nov 11, 2007
  4. Nov 11, 2007 #3
    Ok, so this makes sense, regarding setting up two equations with two unknowns. But if I convert the left side of equation 2 into moles, I still have it equal to grams. I thought about converting the right side into moles, but it doesn't seem like that can actually be done (for example, canceling out the grams with the molecular weight of AgBr + AgCl), & when I tried it, I came up with negative numbers for x & y. Thanks for getting me this far at least.
  5. Nov 11, 2007 #4
    you can't cancel out grams and MM of AgCl, AgBr, they don't have the same units

    MM = g/mol

    both sides of equation 2 are in grams

    MM AgCl =
    mol Cl-

    = g/mol

    gives grams on both sides.

    put the numbers in for MM AgCl and AgBr, to make it look better

    it is not a nice problem because you have your unknowns in the denominator.hope you're good at algebra
  6. Nov 11, 2007 #5
    I broke the second equation down like this

    g AgCl + g AgBr = .9975g

    1 g AgCl x (1 mol AgCl/143.25g AgCl) x (1 mol Cl-/1 mol Ag Cl) = .0069759 mol Cl-

    doing the same for AgBr I got .0053248 mol Br-

    Of course doing this, would leave the other side of the equation in grams.

    Ok, so that doesn't work so well. From what I'm gathering your saying
    187.7 g AgBR/1 mol x 1 mol Br- = g AgBr which I guess would give me 187.7 MM AgBr/1 mol Br-

    For AgCl its 143.15 MM AgCL/1 mol Cl-

    So if I plug those numbers into the equation its 187.7 MM AgBr/1 mol Br- + 143.15 MM AgCl/1 mol Cl- = .9975

    This is where I'm a bit confused. Having the unknown in the denominator wouldn't be a problem, because you could just invert the whole equation, then use whatever method ya like to solve the variables in the two equations (Crammer's Rule is what I usually use for small systems like this one). But does the MM AgBr & MM AgCl every cancel out, or is it just not needed?
  7. Nov 11, 2007 #6
    SORRY! you're right its not in the denominator

    I was muddling my MM = g/mol equn

    this should be right

    equn 2 is 187.77 * y mol Br- + 143.33 * x mol Cl- = .9975 grams

    equn 1 is x + y = .0056568

    let y = .0056568 -x

    then 187.77 (.0056568 - x) + 143.3 * x = .9975 g

    DO be careful with your signs, it looks like x might be neg but you need to bring numbers to the right hand side and keep x on the left (you get neg on both sides which cancel out to give a pos value for x)

    SORRY again about my last post , this one does works out
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook