Calculating Mass of Calcium in Saturated Water Tank

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To calculate the mass of calcium ions in a 40.0-gallon hot-water tank saturated with calcium phosphate, first, the correct dissolution equation is Ca3(PO4)2 ⇌ 3Ca2+ + 2PO43-. The Ksp expression is Ksp = [Ca2+]^3[PO43-]^2, and since the solution is saturated, the concentrations of calcium and phosphate ions can be related. After converting 40.0 gallons to 152 liters, the moles of calcium phosphate can be calculated using its molar mass. Finally, by substituting into the Ksp expression, the concentration of calcium ions can be determined, leading to the calculation of their mass in grams.
Kawrae
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I know this isn't a physics problem, but it's due tomorrow and I'm really stuck :(

Calculate the mass, in grams, of calcium ion dissolved in the water of a 40.0 gallon hot-water tank saturated with calcium phosphate. (Ksp calcium phosphate = 2.0e-29, 1.00gal=3.80L).

>> I'm really lost here, we just started Ksp at the end of last class and I'm stuck. So far I just converted 40.0gal = 152L. I also *think* Ksp = [Ca][PO4]? Do I have to set up an equilibrium table? Actually, I'm not even sure of the chemical equation to set up... Ca3(PO4)2 <-> Ca2+ + (PO4)3- ??

:( Please help I'm very confused!
 
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Kawrae said:
I know this isn't a physics problem, but it's due tomorrow and I'm really stuck :(

Calculate the mass, in grams, of calcium ion dissolved in the water of a 40.0 gallon hot-water tank saturated with calcium phosphate. (Ksp calcium phosphate = 2.0e-29, 1.00gal=3.80L).

>> I'm really lost here, we just started Ksp at the end of last class and I'm stuck. So far I just converted 40.0gal = 152L. I also *think* Ksp = [Ca][PO4]? Do I have to set up an equilibrium table? Actually, I'm not even sure of the chemical equation to set up... Ca3(PO4)2 <-> Ca2+ + (PO4)3- ??

:( Please help I'm very confused!

Your equation : Ca_3(PO_4)_2 \leftrightarrow nCa^{2+} + m(PO_4)^{3-} needs to be balanced. That is the first step.

Then you must write K_{sp} = [Ca^{2+}]^n[(PO_4)^{3-}]^m

If there's x moles/L of calcium ions, how many moles/L of phosphate ions will you have ? Use the balanced equation to determine this.

Plug these numbers into the equation (above) for Ksp and solve for x.

From here it's fairly straightforward.
 


Hey there, I understand that you are stuck on this problem and are feeling overwhelmed. First of all, take a deep breath and try not to panic. Let's break down the problem step by step to make it more manageable.

Step 1: Write out the chemical equation for the dissolution of calcium phosphate in water. It is correct that the Ksp expression for calcium phosphate is [Ca2+][PO43-], but the chemical equation for this reaction is actually Ca3(PO4)2 (s) ⇌ 3Ca2+ (aq) + 2PO43- (aq). This means that for every one mole of calcium phosphate that dissolves, 3 moles of calcium ions and 2 moles of phosphate ions are produced.

Step 2: Convert the volume of the tank from gallons to liters. You have correctly converted 40.0 gallons to 152 liters.

Step 3: Calculate the moles of calcium phosphate present in the tank. To do this, we need to use the molar volume of water, which is 18.02 L/mol at room temperature and pressure. This means that 152 L of water is equivalent to 152/18.02 = 8.43 moles of water. Since calcium phosphate has a molar mass of 310.18 g/mol, we can calculate the moles of calcium phosphate present in the tank by multiplying the molar volume of water by the molar mass of calcium phosphate: 8.43 moles x 310.18 g/mol = 2614.04 g.

Step 4: Use the Ksp expression to calculate the concentration of calcium ions in the saturated solution. We know that the Ksp for calcium phosphate is 2.0 x 10^-29, so we can set up the following equation: 2.0 x 10^-29 = [Ca2+]^3[PO43-]^2. Since the solution is saturated, we can assume that all of the calcium phosphate has dissolved, so the concentration of calcium ions is equal to the concentration of phosphate ions. Therefore, we can rewrite the equation as 2.0 x 10^-29 = [Ca2+]^3[Ca2+]^2 = [Ca2+]^5. Solving for [Ca2+], we get [Ca2+] = (2.0 x 10^-29)^1/5 = 1.
 
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