(c) If 55.8 mL of a BaCl2 solution
is needed to precipitate all t...

Question

(c) If 55.8 mL of a BaCl2 solution
is needed to precipitate all the sulfate ion in a 752-mg sample
of Na2SO4, what is the molarity of the BaCl2 solution?

Accepted Answer

Hey everyone. Our question here states that one strontium fluoride reacts with sodium carbonate. Solid strontium carbonate is formed. Were asked to calculate the polarity of our strontium fluoride solution 1st. Let's go ahead and write out our reaction. So we were told that we have sodium carbonate and this is Aquarius and we react it with strontium fluoride, which is also Aquarius. And when these two react, we end up with solid strontium chrome eight and sodium fluoride. Now, before we move forward, let's go ahead and balance this out. As we could see in our react inside, we have two of sodium. So we'll need to add a coefficient of two prior to our sodium fluoride. Looking at everything else, we can already see that it is completely balanced out. And we can go ahead and move on with our question. Starting off with our 642 mg of sodium carbonate. We're going to convert this into grams and we know that we have 10 to the third milligrams per one g. Next, we're going to take sodium carbonates, Mueller mass which is 105.99 g or one mole. Now we're going to look at our multiple ratios between sodium carbonate, which is one more hmm, one more of strontium fluoride and this will get us a value of 6.57 times 10 to the negative third mole of strontium fluoride. And since we want the polarity of our strontium fluoride. We're going to take that value of 6.57 times 10 to the negative third mole And divide it by 43.2 ml. And since we want this in, leaders were going to use dimensional analysis and we know that we have 10 to the third middle leaders per one leader. And this is going to get us to a more clarity of 0. of strontium fluoride. So I hope this made sense and let us know if you have any questions.

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