Assume that you dissolve 10.0 g of a mixture of NaOH and Ba(OH)2 ...

Question

Assume that you dissolve 10.0 g of a mixture of NaOH and Ba(OH)2 in 250.0 mL of water and titrate with 1.50 M hydrochloric acid. The titration is complete after 108.9 mL of the acid has been added. What is the mass in grams of each substance in the mixture?

Accepted Answer

Hello everyone today. We have the following problem. Consider dissolving 7.5 g of a potassium hydroxide and calcium hydroxide mixture. And 250 mL of water before tai trading with 1.75 Mueller of hydrochloric acid. After adding 100 and two mL of acid, that tight rations finished. How much weight does each substance in the mixture contribute in grams. So the first thing that you want to make note is that if we let X equal our mass of potassium hydroxide, then the total mass. So if X equals the massive potassium hydroxide, then the mass of our calcium hydroxide Will equal 7.5 or the total mass 7.50 -1. And this is to simplify for variables. The second thing you want to do is he want to find moles so he wants to find the moles of hydrochloric acid. And so to do that, we're going to start with our given. So what do we have of hydrochloric acid? We have our 102 middle leaders. So we take our 102 middle leaders. We of course convert this into leaders by using the conversion factor that one mil leaders equal to 10 to the negative third leaders. We then multiply by our polarity. And so the maturity was given to us by 1.75 molar and molar it is in terms of moles per one leader. So we can just substitute that in there for moles for polarity. Our units for leaders and our units from middle leaders canceled out. And we're left with 0.1785 moles of our hydrochloric acid. And so essentially what we want to do next is we want to find our mass of X. R. Mass of potassium hydroxide first. And so how can we go about doing that? Well, the first thing that we can do is we can set up an intricate equations up. So we have X. Here we are going to multiply that By one mole of potassium hydroxide Over its molar mass which is 56.11 g. This is of course according to the periodic table. And then what we're going to do next is we're going to add two Times the total number of weight that we have in grams over our molar mass of calcium hydroxide which is 74.09 g. And this is once again according to the periodic table. And so why do we have that to there? That too is simply to represent that We have two of our reactant are potassium hydroxide and calcium hydroxide. And so the next thing that we want to do is we want to subtract by once again our two X for potassium hydroxide over our secondary Molar mass which is 74.9 g of calcium hydroxide. And we want to equal this to the moles of hydrochloric acid that we found. So essentially we're going to be wanting to solve for X. And after calculations we arrive with X is equal to 2.6119 grams. And so now we can put this together X represented our mass of potassium hydroxide. So our massive potassium hydroxide is 2.6119g. And our mass of calcium Hydroxide is equal to the total mass or 7.50 g -2.6119g to give us 4. 881 g of calcium hydroxide. And so with that we have solved this question. Overall, I hope this helped, and until next time.

Watch next