How many moles of ions are in 1.75 mol of K₂SO₄?

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Welcome back, everyone. A sample contains 3.25 moles of rubidium oxalate. How many moles of ions are present in the sample? We are going to recognize that rubidium oxalate consists of rubidium which is R B and it is bonded to our oxalate anion. So we've got R B sub two C 204 recall that oxalate is a poly atomic anion. So we've got the combination of rabi a metal plus a poly atomic, an ion. And this therefore means that we've got an ionic compound. So this is ultimately therefore going to be a salt. So we'll say that this is solid and we should recall that rubidium oxalate is actually highly insoluble in water. So it's going to only partially dissociate. So it's going to be an equilibrium. And as a result, we would form two moles of our rubidium plus one ions along with one mole of our oxalate two minus anion. So we need to determine the moles of ions present in the sample. First, let's begin with the total moles n of our rubidium plus one ions. So we're going to start with the 3.25 moles of rubidium oxalate R B two C two oh four. And we will multiply to go from moles of rubidium oxalate in the denominator two moles of rubidium. I'm sorry, this should be rubidium ions in the new. So based on our equation, our dissociation equation that we've written out as an equilibrium, we would observe that we've got for one mole of rubidium oxalate dissolved approximately two moles of rubidium ions and equilibrium as a product. So we've got a 1 to 2 molar ratio. So, canceling out moles of rubidium oxalate, we are left with moles of our rubidium plus one cat ions. And we would find our moles of rubidium cats equal to a value of 6.50 moles of rubidium plus one ions. So now for the moles of our oxalate anions, we're going to again begin with the 3.25 moles of rubidium oxalate and multiply to go from a molar ratio. Where in our denominator, we have moles of rubidium oxalate, dissolved pear moles of oxalate dissociated in the numerator, oxalate ions dissociated. And so using again our molar ratio from our equation, we've got for one mole of rubidium oxalate, dissolved one mole of oxalate ions dissociated. And so this will then allow us to determine that we've got 3.25 moles of oxalate anions, C 2042 minus in solution. So now with the moles of each of our ions, we then want to calculate the total moles of ions. And so we have n total ions for total moles of ions. And we would then take our moles of rubidium plus one ions added to our moles of oxalate anions dissociated. So, taking the sum, we would have 6.50 moles of our rubidium plus one ions added to 3.25 moles of our oxalate two minus anions. And in total, that would give us our total moles of ions being a value of 9.75 moles of ions. So for the ions in total present in our sample of rubidium oxalate, our final answer is going to be 9.7 moles. I hope that this made sense and let us know if you have any questions.

How many moles of ions are in 1.75 mol of K₂SO₄?

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