A 500.0 mL sample of saturated nickel(II) phosphate solution was analyzed spectrophotometrically and found
to contain 1.88 × 10-5 g of nickel.
(a) Determine the molarity of the nickel(II) phosphate solution.
(b) Calculate the solubility product constant, Ksp for nickel(II) phosphate
a. 2.1*10^-7
b.4.85*106-32
these are correct
I'm assuming that the problem below (which I did in detail) will help you work this one.
a. 1.6
b.7.8*10^-6
The above answer is wrong.
To determine the molarity of the nickel(II) phosphate solution, we need to use the given information:
Volume of the solution (V) = 500.0 mL = 0.5000 L
Mass of nickel (m) = 1.88 × 10^(-5) g
(a) To find the molarity (M) of the nickel(II) phosphate solution, we use the formula:
Molarity (M) = Number of moles (n) / Volume (V)
First, we need to calculate the number of moles of nickel using its molar mass. The molar mass of nickel (Ni) is 58.69 g/mol. Therefore,
Number of moles of nickel (n) = Mass of nickel (m) / Molar mass of nickel (Mn)
= 1.88 × 10^(-5) g / 58.69 g/mol
Now, we can calculate the molarity using the formula:
Molarity (M) = n / V
= (1.88 × 10^(-5) g / 58.69 g/mol) / 0.5000 L
Substituting the values and solving the equation will give us the molarity of the nickel(II) phosphate solution.
(b) To calculate the solubility product constant (Ksp) for nickel(II) phosphate, we need to use the chemical equation for the dissociation of nickel(II) phosphate:
2Ni3(PO4)2(s) ⇌ 3Ni2+(aq) + 2(PO4)3-(aq)
The solubility product constant (Ksp) expression for this equation is:
Ksp = [Ni2+]^3 * [(PO4)3-]^2
Since nickel(II) phosphate is a sparingly soluble salt, we can assume that the concentration of nickel and phosphate ions in the saturated solution is equal to the solubility (S) of the salt.
Using the molarity (M) we calculated in part (a), we can assume the concentrations of [Ni2+] and [(PO4)3-] are equal to the solubility (S) of the salt.
Thus, we have:
Ksp = S^3 * S^2
= S^5
We need to find the value of S, which is the solubility of nickel(II) phosphate.
To do that, we can use the given information about the mass of nickel (1.88 × 10^(-5) g) and the molar mass of nickel(II) phosphate (Ni3(PO4)2).
Molar mass of nickel(II) phosphate (Mp) = Mass of nickel(II) phosphate / Number of moles of nickel(II) phosphate
Since the chemical equation shows that there are 3 moles of nickel atoms in 1 mole of nickel(II) phosphate, we can convert the mass of nickel to moles:
Number of moles of nickel(II) phosphate = Mass of nickel / (Molar mass of nickel(II) phosphate * 3)
Now, we can calculate the solubility (S) using the equation:
Solubility (S) = Number of moles of nickel(II) phosphate / Volume (V)
Thus, S = (Mass of nickel / (Molar mass of nickel(II) phosphate * 3)) / Volume
Substituting the given values and solving the equation will give us the solubility (S) of nickel(II) phosphate.
Finally, to calculate the solubility product constant (Ksp), substitute the solubility (S) into the Ksp expression:
Ksp = S^5
Solve the equation to find the value of Ksp for nickel(II) phosphate.