[aaazureee]

When concentration of Ca2+ ions in water is greater than 10^-5 M (mol dm-3), an insoluble "scum" would be formed with soap. To prevent scum forming with soap, what minimum mass of sodium carbonate, Na2CO3 should be added to 1 dm3 of water? (Ksp of CaCO3 = 5x10^-9)?
Im quite confused how to do this problem I have tried look up Ksp, solubility and common ion effect but i cant figure out the answer?

[Borek]

Hint: what is the minimum concentration of CO₃^2- that guarantees the concentration of Ca²⁺ to be lower or equal 10^-5 M?

[aaazureee]

So I have tried to solve this problem:
CaCO3-> Ca2+ + CO3 2-
Ksp = [Ca2+] x [CO3 2-]
5x10-9 = [10^-5] x [CO3 2-]
=> [CO3 2-] = 5 x 10^-4
=> mass of Na2Co3 = 5x10^-4 x 106 = 0.053g
Is that correct?

[Borek]

Looks OK to me.

Disclaimer - in general this is not an exact answer, as carbonate reacts with water

CO₃^2- + H₂O HCO₃^- + OH^-

so its concentration is lower than expected. But I doubt you were expected to take it into account.

[aaazureee]

I have checked the answer and its 0.53g. I cant figure it out somehow lmao

[Borek]

Must be a mistake, I got 0.053 g just like you did.