Question

A sample of solid Y(OH)3 was stirred in water at a certain temperature until the solution contained as much dissolved Y(OH)3 as it could hold. A 550-mL sample of this solution was withdrawn and titrated with 7.74e-05 M HI. It required 50.1 mL of the acid solution for neutralization. What is the solubility of Y(OH)3 in water, at the experimental temperature, in grams of Y(OH)3 per liter of solution?

Answer 1

Solubility of Y(OH)3 = 3.29 * 10^-4 g/ L

Step-by-step explanation:

Step 1: Data given

Volume of Y(OH)3 = 550 mL = 0.550 L

Molarity of HI =7.74 * 10^-5 M

Volume of HI = 50.1 mL = 0.0501 L

Step 2: Calculate the moles of HI

Moles HI = molarity * volume

Moles HI = 7.74 * 10^-5 M * 0.0501 L

Moles HI = 0.00000387774

Step 3: Calculate moles Y(OH)3 needed

For 1 mol Y(OH)3 we need 3 moles HI

For 0.00000387774 moles HI we need 0.00000387774/3 = 0.00000129258 mol

Step 4: Calculate mass Y(OH)3

Mass Y(OH)3 = moles * molar mass

Mass Y(OH)3 = 0.00000129258 mol * 139.93 g/mol

Mass Y(OH)3 = 0.00018087 grams

Step 5: Calculate the solubility of Y(OH)3 per liter

The solubility of Y(OH)3 is 0.00018087 grams in 0.55L

Solubility of Y(OH)3 = 1000/550 * 0.00018087 grams

Solubility of Y(OH)3 = 3.29 * 10^-4 g/ L