What is the molarity of each ion in a solution prepared by dissol...

Question

What is the molarity of each ion in a solution prepared by dissolving 0.550 g of Na2SO4, 1.188 g of Na3PO4, and 0.223 g of Li2SO4 in water and diluting to a volume of 100.00 mL?

Accepted Answer

Hello everyone in this video we want to calculate for the molar concentrations of each ion. That's using these regions here, that's dissolved in water and then diluted to a 250 millimeter volume. So these molecules are all a quiz and they dissociate into its ionic forms when it's in water. So let's go ahead and write out these reactions per molecule. So the first free agent that we're starting off with is C. A. S. 04 That associates into our calcium two plus and S. 042 minus Second one we have is K two C. 03 that associates into two moles of our potassium plus Canadians and one mole of C. +032 minus. And last one fear is K. Two S. 04. So very similar to the reaction above. We have two moles of potassium plus cannons And one mole of s. 0. 4 2 -. Alright so let's go ahead and calculate now for the polarity of each molecule in solution. Let's go ahead and recall that capital M. Which similarity or units of concentration is able to moles of solute over the leaders of solution. So coupling for the leaders of solution here Given that I sell it to 250 million years. So that's our unit or volume of our solution. So we just want to convert this into leaders. Then We just do a direct conversion for every ml. We have one leaders. The units of military council's putting this into my calculator, get the value of 0.250 L. So it's going to be our denominator for all of our calculations here. Alright, so now I'm gonna go ahead and sell for my concentration of our first regent that c. s. 0. 4. So given that we have 1.38 g of our region, here we go ahead. And use the lower massiveness. Give us moles Which is what we want for our numerator. Alright? So for every one mole of this product we have 136.14 g of this. So you see now that the gram unit will cancel. So once we put this in the calculator we get the value of we'll just put the whole thing that we see here on display that 0. units being moles of RCAS 04. So now we have our new murder. Let's go ahead and divide this by our denominator that we say about 0.250 L. So once we do that we get the value of 0.405 capital M. For morality of C A. S. 04. Now doing it zack same calculation for our second one here. So we have K two C 03. Now we're starting off with 2. g of our regent. Again, we're just gonna go ahead and use the molar mass to guess in terms of moles. So for every one mole of this region we have .205g of this region. Let me see. The gram unit will cancel. So now we're putting this into the calculator again. I wrote the whole value. I see on the display we have 0. moles of K two C. 03. Again I go ahead and divide this by our predetermined The dominator to be 0.250 L. Once you put that into a calculator we get the value of our concentration to be 0.677 moller for K two C. 03. Last molecule that we need to go ahead and calculate is for the concentration of K two S. 04. This time we're starting off with 1.91 g of this reagent. And the molar mass for this is that for everyone more We have a 174. g game. You can see the units of Graham canceling. Alright. We can then get the value. Once we put it into the calculator of 0.1096069. That's in units of moles of our K two S. 04, Dividing it by our predetermined denominator of 0.250 L. We then get the value of 0.04 Polarity of K two s. 04. Alright now we can go ahead and actually calculate for each ion. So let's go ahead and scroll down just a little bit. Alright so for the concentration of R. C. A. Two plus. So you see that we only have one source of calcium here from C. A. S. 04. It's a 1 to 1 ratio. So we'll just have this much of the concentration of our calcium two plus 0.405 polarity of C. A. Two plus. Next we calculate for the concentration of K. Plus. So we see from our reactions that it's always a two mole Or 2-1 ratio. So we'll multiply whatever number we have for these values by two. So for our first one it's two times 0.677. That's from our K. Two C. 03 here And then we'll add that with two times 0.0438. And that's just for the K two s. 04. So unless we do so we get the value of 0. moller for our K. Plus. That would do the same thing For our s. 0. 4 2 -. So then we have again two different sources, we have it from the first region and then the last one. So everything here is a 1-1 ratio. So add this number directly with again we're just one mole here. So we'll just use this value. So we have value of 0.405 plus 0.438. We get the value or sum of 0.834 moles similarity, sorry, of S 042 minus. And lastly, we have to solve for the concentration of the c minus ion. So that is just from this reagent here and that's just a one ratio. So we just used this value That's 0.0677 polarity of our CO 32 -1. So then these are going to be our concentrations for each ion and the regions that are used.

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