An 8.65-g sample of an unknown group 2 metal hydroxide is dissolved in 85.0 mL of water. An acid–base indicator is added and the resulting solution is titrated with 2.50 M HCl(aq) solution. The indicator changes color, signaling that the equivalence point has been reached, after 56.9 mL of the hydrochloric acid solution has been added. (b) What is the identity of the metal cation: Ca2+, Sr2+, or Ba2+?

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Hi everyone for this problem, we're told that a base with the formula of M. O. H. And a mass of 5.57 g was dissolved in 100 mL of water. The resulting solution was titrate, ID with a 1.65 molar hydrochloric acid solution. A volume of 60.2 mil. Leaders of the hydrochloric acid solution was needed to reach the equivalence point, determine the identity of em. Okay, so for this problem, let's let m represent our metal. So we'll say that M is our metal. And we want to determine the identity of this. And so we need to write a reaction. So we know we have an acid and a base and whenever we have an acid and a base react, it forms salt and water. So our reaction we know that our base is M. O. H. We're told that our acid is hydrochloric acid and this is going to form a salt and water. So our salt is going to be M. C L. Okay? And we have water. Okay, so we can see here that are multiple ratio for our acid and our base is one. So we have a one to mole ratio and we can start off by the first thing we need to do is determine our moles of base. And we're going to start off with the volume of acid we were given. So we're told that we have 60. milliliters of acid. Okay, so that's right here. So we're going to start there. So we're going to go from volume of acid, two moles of base. So 60.2 mL of hydrochloric acid. We need to go from milliliters to leaders in order for us to use our mill arat E conversion. So in one millimeter we have 10 to the negative three leaders. Okay, so now our milliliters cancel and we're left with leaders and we want to go like we said from volume of acid, two moles of base. And the problem they gave us the mill arat E of hydrochloric acid, they tell us it's 1.65 molar. And recall that malaria t means moles over leader. So that means we have 1.65 moles of hydrochloric acid for every one liter of solution. So we can use that as a conversion here. So 1.65 moles of hydrochloric acid for every one leader of solution. So now our leaders cancel and we're left with moles of hydrochloric acid. Our end goal is to get two moles of base molds of M. O. H. So now we can use our multiple ratio between our acid and our base to convert between the two. And we said that it's a 1 to 1 ratio. So we have one more Whole of hydrochloric acid for every one mole of base. And that's why we wrote our wrote out our reaction. Okay, so we have our moles of hydrochloric acid cancel. And we're left with moles of base. So let's go ahead and do this calculation. And when we do we get 0. moles of base. Okay, so we started off with our acid volume of acid and we figured out our moles of base. So now that we know are moles of base, we can calculate the molar mass of our base. Okay, so our moller mass of our base. Remember molar mass is equal to mass Over moles. So we know what our masses because it was given to us in the problem we were told we have 5.57 g of base and we just calculated our molds of base and we set at 0. moles of base. Okay, so now our molar mass of our base is 56.8 g per mole. The problem asked us to determine the identity of em are metal. So what we can do now is because we know that we have hydrogen and oxygen in our base because our base is M. O. H. And we have an oxygen and a hydrogen. What we can do is subtract the mass of hydrogen and oxygen from our base here. So we have a we have a metal, we have oxygen and we have a hydrogen. We know the molar mass of our our base as a whole. So now we can subtract the molar mass of hydrogen and oxygen to determine the identity. So let's do that. So we have our our molar mass of our M. Is going to equal our total molar mass of our base, which we calculated is 56.808 g per mole minus Our mass for hydrogen and oxygen. So our mass for hydrogen is 1.01 g per mole, And our molar mass of oxygen is g per mole. Okay, so I'll make a note above this is hydrogen and this is oxygen. So now we will get our molar mass Of our mettle is going to equal 39.1 g per mole. So now we can look at our periodic table to see what is most closely what is the closest to this molar mass. And when we do we see that it is potassium. Okay. Which has the symbol K. So this is going to be the identity of M. And that is our final answer. That's the end of this problem. I hope this was helpful.

Verified Solution

Key Concepts

Titration and Equivalence Point

Stoichiometry of Acid-Base Reactions

Group 2 Metal Hydroxides