A 100.0 mL solution containing aqueous HCl and HBr was titrated w...

Question

A 100.0 mL solution containing aqueous HCl and HBr was titrated with 0.1235 M NaOH. The volume of base required to neutralize the acid was 47.14 mL. Aqueous AgNO3 was then added to precipitate the Cl-and Br-ions as AgCl and AgBr. The mass of the silver halides obtained was 0.9974 g. What are the molarities of the HCl and HBr in the original solution?

Accepted Answer

15 molar of potassium hydroxide was used to titrate a 150 mL solution containing aqueous hcl hbr 17. mL of base was needed to neutralize the acid, the chloride and bromide ions within precipitated as copper chloride and copper bromide. By the addition of acquis copper nitrate. The copper highlights produced had a mass of 0.89 34 g. Hcl hbr polarities in the initial solution. So the first thing that we need to do is we need to find moles of H plus and so we can take into account that the moles in our hcl plus the moles of our HPR equal the total number of protons that we have. And so we start off with our Volume 17. ml. We use the conversion factor that one middle leader is equal to 10 to the negative third leaders. It will be multiplied by the polarity which is going to be 15 point oh moles per one liter are units of liters and milliliters cancel out and we end up with 0. 6-5 moles of each plus. We then have to make note of our total mass. And so we're gonna take our mass of our copper chloride and our mass of copper bromine. And this is going to equal 0. g. So we're gonna hold on to this. These two numbers that we solved for. We also need to make note that the moles of our copper are equal to the moles that we have of our H plus protons. Next we must find our mass of copper and grams. And so how do we do that? Well, we know that we have a we know that we have the mass of copper and the mass of our bromine. And this is going to equal 16. g of copper. We then take the mass that we have of chloride plus the mass of our bromine And that's going to equal 34. -16.6808 g to give us 17.6-2 g of copper. For this next step, we're gonna let X equal the most of chlorine. And why will equal the moles of borrowing. And so for copper, We already said that we have 16.68 grams of copper that's going to equal X plus y are moles of copper plus our moles of roaming Times. Our Mueller massive copper, which is 63.546 g. And when we, when we rearrange this equation, we're going to end with X is equal to 16.68 over 63.546 minus y. And so we're gonna solve simultaneously for X and Y. So when we end up doing that, we are going to get a value of X, Our value of X is going to equal 0.075 moles, I will be for chlorine and why for browning is going to equal 0.18707 moles. And so therefore our concentrations for our last step for hcl is going to be the molds that we found. 0.075 Malls over our volume of 0.15 liters. Since we had 150 mL, we just simply divided that by 1000 to get leaders and to our concentration for Hcl be 10000.5 molar. And then our concentration for our HBR Is going to be 0. 707 moles over 0.50 L to give us 1.25 moller as our final answer. And with that we've answered the question overall, I hope that this helped. And until next time.

Watch next