A solution of 105.0 mL of 0.300 M NaOH is mixed with a
solution o...

Question

A solution of 105.0 mL of 0.300 M NaOH is mixed with a
solution of 150.0 mL of 0.060 M AlCl3. (e) What is the concentration
of Cl- that remains in solution?

Accepted Answer

Hi everyone here we have a question telling us that the solution is created by mixing mL of 0.320 moller barium bromide and 130 mL of 0.28 to 18 moller magnesium sulfate. And our question is how much bro mean? And polarity is left in the solution. So first we need a balanced equation. So we have barry and bromide and we have magnesium sulfate. This is a double displacement reaction. So are very um with a charge of two plus is going to go with our sulfate with a charge of two plus or two minus. So those are going to cancel out and form barium sulfate. Our magnesium is going to go with our bromine. Magnesium has a charge of two plus bromine has a charge of one minus. So we get magnesium and to browning so that's forming barium sulfate plus magnesium bromide. And if we look, we see that this solution or this reaction is already balanced. So we can go on to our next step which is to calculate the amount of the reactant. So we have milliliters And we're going to change that to leaders. Sometimes one leader Divided by ml times 0. moles of barry and bromide over one leader of barium bromide and that's because polarity is moles over leader and then we're going to Calculate that and it gives us 0. moles of barium bromide. And if we look at our dimensional analysis are male leaders are canceling out and our leaders are canceling out, leaving us with moles which is exactly what we want. So now we're going to do the same thing with our 130 ml. So we're gonna multiply that by one leader over 1000 ml to convert it to leaders. And then we're going to multiply that by 0. moles, magnesium sulfate over one L of magnesium sulfate. And our middle leaders are canceling out and our leaders are canceling out and that gives us 0. moles a magnesium sulfate. Now we need to determine the limiting reactant. So we are going to start with barium bromide and go to barium sulfate. So we have 0. moles a barium bromide times one mole a barium sulfate. And that's from our balanced reaction over one more a barium bromide And our moles of barium bromide are canceling out, giving us 0. moles of barium sulfate. And now we have 0. for moles of magnesium sulfate, times one more of barium sulfate. And this is our multiple ratio From our balanced equation over one mole of magnesium sulfate And that equals 0. moles of barium sulfate. So we look and see which one we have less of and we get less when we use the magnesium sulfate. So it is our limiting reactant. Next we calculate the barium bromide consumed and remaining. So we have barry and bromide consumed Equals 0.02834 moles of barium sulfate. And we got that by calculating the molds of magnesium sulfate to barium sulfate. And we're going to multiply by our multiple ratio. So times one mole barry and bromide over one mole barium sulfate and that's from our balanced equation And that equals 0.02834 moles. So sarah barium bromide remaining equal 0.0384 moles. Because that's what we began with -0. moles because that is what was consumed And that equals 0.01006 moles. So now we're going to take our total volume Which is 0.120 L plus 0.130 L. And that equals 0.25 L. So we have 0. moles of barry and bromide. Times two moles of romaine over one mole a barium bromide And that equals 0. 2012 moles of bromide. And now we need to calculate our mobile charity. So our mill arat E of bromide is 0. moles A bromide and if we remember correctly polarity is moles over L. So we're going to divide that by the leaders of bromide, which we determined was 0.25 l And that equals 0. polarity. And that is our final answer. Thank you for watching. Bye.

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