How many total moles of ions are released when the following sample dissolves completely in water?
31.3 g of Ba(OH)2 · 8H2O
moles Ba(OH)2.8H2O = grams/molar mass = 31.3/315.467 = 0.0992
So you have 1 mols Ba^2+ for every 1 mol Ba(OH)2.8H2O; therefore you must have 0.0992 mols Ba^2+. Following up you will have
2 x that for OH^- ions. In addition you will have an adidtional 8 x that for H2O molecules. I assume you're not counting the small amount of H^+ and OH^- that come from the ionization of the H2O molecules.
To determine the number of moles of ions released when 31.3 g of Ba(OH)2 · 8H2O dissolves completely in water, we need to consider the dissociation of the compound in water.
Ba(OH)2 dissociates into Ba^2+ ions and OH^- ions.
The coefficient in front of Ba(OH)2 in the chemical formula is 1, indicating that there is one mole of Ba(OH)2 in the given sample.
To calculate the moles, we need to find the molar mass of Ba(OH)2 · 8H2O:
- The molar mass of Ba is 137.33 g/mol.
- The molar mass of O is 16.00 g/mol.
- The molar mass of H is 1.01 g/mol.
Thus, the molar mass of water (H2O) is:
2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol.
Since there are 8 H2O molecules in the compound, the total molar mass of Ba(OH)2 · 8H2O is:
(137.33 g/mol) + 2(16.00 g/mol) + 8(18.02 g/mol) = 315.28 g/mol.
Now we can calculate the number of moles of Ba(OH)2 · 8H2O:
Number of moles = mass / molar mass
Number of moles = 31.3 g / 315.28 g/mol.
By dividing 31.3 g by 315.28 g/mol, we find that there are approximately 0.099 moles of Ba(OH)2 · 8H2O in the given sample.
Since each formula unit of Ba(OH)2 generates two ions (Ba^2+ and 2OH^-), we multiply the number of moles by the coefficient (2) to get the total number of moles of ions released:
Total moles of ions = 0.099 moles * 2
Therefore, when 31.3 g of Ba(OH)2 · 8H2O dissolves completely in water, approximately 0.198 moles of ions are released.