A 1.268 g sample of a metal carbonate (MCO3) was treated with 100.00 mL of 0.1083 M sulfuric acid (H2SO4), yielding CO2 gas and an aqueous solution of the metal sulfate (MSO4). The solution was boiled to remove all the dissolved CO2 and was then titrated with 0.1241 M NaOH. A 71.02 mL volume of NaOH was required to neutralize the excess H2SO4. (a) What is the identity of the metal M?

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Hello everyone. So this question we're being told that an agreed solution of a male nitrate. So M. N. 03 and cl two gas was produced when a sample of a metal chloride weighing this many grams was treated with this amount of nitric acid. So the solution was boiled to get rid of all the dissolved cl two gas. The solution was then tie treated with a 0.2 molar K. O. H. Here trying to determine the identity of the metal M. If this much of K. O. H. was required to neutralize the excess h. n. 0. 3. Alright so first we need to go ahead and determine the number of moles of H. And 03 and K. O. H. We're gonna go ahead and use this equation right here which is N. Equals to M. Times V. So M. Is our concentration. That's the maturity and V. Is over our volume. So we know that this capital M. For molar itty the units for that is moles over leaders. And the reason why we multiply it with RV here, the volume. It can go ahead and cancel out with the leaders on the bottom of this here. Alright so let's first calculate the volume For our h. n. 0. 3. So this will be equal to so we're given 92. male leaders. We're gonna first convert this into leaders. So for every one leader we have one times 10 to the third mila leaders. So we can see here that the middle leaders unit will cancel leaving us with the units of leaders. So once I put this into my calculator I get my volume for H. And 03 is equal to 0.9 to seven leaders. Now continuing our calculations we go ahead and now solve for R. N. So the moles of our H. And 03. Again this is equal to the polarity of the H. I know three as well as our volume of the H. And oh three. So putting in the numbers that we already have. So for this one more per leader this is the m similarity to multiply this by the volume. And he said this is going to be 0.9 to seven liters. So we can see here that the leaders will cancel giving us just are moles which is exactly what we want. So putting this into my calculator I see that we get 0.927 moles of R H. N. 03. Alright so now in a different color I can go ahead and sell for my volume of my K. O. H. So again we're given 196 mL of this solution. We want to go and convert this into leaders. So again for everyone leader that we have we have one times 10 to the third milliliters. Again we see that the male leaders can cancel. So once I put this into my calculator I get to my volume is equal to 0.196 L. Okay same thing now we're selling for the end. So the moles but this time of R K. O. H. So it's the same exact equation that we did here except now regarding our K. O. H. So I won't rewrite to this occasion. But just now we're using the same exact one. So for the modularity so the concentration of my K. O. H. We have 0.200 Moles for everyone leader. We want to multiply this by the volume that we just solve for. So that is 0.196 L. You can see that the leaders will cancel. Alright so putting this into my calculator, I get that the molds of my K. O. H. Is equal to 0.392 units being moles of R K. O. H. So the equation is that when we combine the H. And oh three which is acquis and R K. O. H. Which again is acquis. This will yield the K. N. 03 As well as H20. And of course that's in its liquid state. So we see here that the two reactions are going to be a 1-1 ratio. Alright, so here's our next step, what we're gonna do is determine the number of moles of excess H. N. 03 that's neutralized by R K. O. H. So we'll go ahead and do this with a different color. So again we're solving for the moles of excess each and three. So we have this 0. moles of K. O. H. So we're gonna do a multiple ratio here already said earlier that for every one mole of H and 03 we have one mole of R K. O H. And it's determined by our chemical reaction. Where he said that the two reactant are a 1 to 1 ratio. So once you put this into the calculator, we see that we get 0.392 moles of access will actually I will simplify to X. S. H. And 03. All right. So now with this information we can go ahead and determine the number of moles of H and 03 that reacted with M. C. L. By subtracting the moles of excess H and 03 from the initial number of moles of H and 03. I'm gonna go ahead, scroll down for more space here. All right, so for the moles of my H. N. 03 that's reacted with our M. C. L. This is equal to 0.0927 moles minus. R 0.392 moles. Putting this into calculator. I get that this difference is 0. moles. Again, it's the moles of our cl minus and R M. C. L. And this gives us the 0.5 35 moles of cl minus. Alright. And some information that we should know here is that the molecular weight of our c l minus is 35.453 rams per mole. So we just use the molecular weight from the pr table of cl Alright, so now we can go ahead and determine the mass of cl minus by using our molecular weight. The equation that will use is just that this mass is equal to the ends of most times molecular weight. Alright, so again, we're calculating here for the mass of our cl minus. Again, we're using the number of moles. So that 0.535 moles of cl minus. We're gonna go ahead and multiply this with a molecular weight which we just wrote above that being 35.453 g for every mole of cl or cl minus. In this case we can see here then that the moles will cancel leaving us with just the units of grams which is exactly what we want for our molecular weight or the mass. Rather so putting this into my calculator, we get 1.8967 g of cl minus. Now let's go ahead and determine the mass of the metal. M by subtracting the mass of cl minus from the initial mass of the sample. So again, I'm gonna go and scroll down here for more space. So the mass of our m metal is equal to 7.668 g minus the 1.8967. Which is of our cl minus here. So once I put that into my calculator, I get the difference of 5.7713 g. Now calculating for the molecular weight of this unknown metal. M we should know that the molecular weight is equal to the mass over the number of moles. So here, just using that equation, we also know that the moles of the metal equals to the moles of our coin. And that's just 0.0535 moles. So again using this equation here, but gold on top is our mass and that is 5. g. And we said the N. Is equal to the end of our C. L. And that again is 0.535 moles. So once you put that into a calculator, we see that the molecular weight of the M metal is equal to one oh 7.87 g Permal. Now using this molecular weight can go ahead and check out the pureed table. You see which metal lines up with this answer here. I can see from my parent table that this mass here closely relates to our metal, silver with the element symbol of a G. So we can say that our metal M. Is going to be silver. And this right here is going to meet my final answer for this problem

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Key Concepts

Stoichiometry

Acid-Base Reactions

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