Question

You prepare a solid salt with formula X(OH)3. You wash the solid, redissolved most but not all of it in water and pipetted a 4.00 mL sample of the solution. This sample was titrated with a 0.0100 M HCl. The sample required 7.33 mL of the hydrochloric acid to turn phenolphthalein to clear. What is the Ksp for this salt?

Answer 1

Neutralisation equation is,

X(OH)3 (aq) + 3 HCl (aq) -----------> XCl3 (aq) + 3 H2O (l)

Neutralisation formula,

M1 V1 / n1 = M2 V2 / n2

M1 X 4.00 / 1 = 0.0100 X 7.33 / 3

M1 = 0.0100 X 7.33 / ( 3 X 4.00 )

M1 = Concentration of X(OH)3 = 0.00611 M

Solubility equation is,

X(OH)3 (s) $\small \rightleftharpoons$ X^3+ (aq) + 3 OH^- (aq)

[X3+] = [X(OH)3] = 0.00611 M

[OH^-] = 3 X [X(OH)3] = 3 X 0.00611 = 0.0183 M

Solubility product expression can be written as,

Ksp = [X^3+][OH-]^3

Ksp = ( 0.00611 ) * ( 0.0183 )^3

Ksp = 3.74 X 10^-8