[Lurifax]

I'm working on a problem, but I'm not quite sure how to get started on this

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Problem:

Hard water often contains dissolved Ca²⁺ and Mg²⁺ ions. One way to soften water is to add sodium carbonate, Na₂CO₃.
The carbonate ion (CO^2-₃) forms insoluble percipitates with calcium and magnesium ions, removing them from solution.

Suppose that a solution is 0.050 M in calcium and 0.085 M magnesium nitrate.
What mass of sodium carbonate would have to be added to 1.5 L of this solution to eliminate the hard water ions?
Assume complete reactions.

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molar mass of Na₂CO₃: 106 g/mol

I have equations such as:

MM(molar mass) = m(mass) / n(moles)

and

M(molarity) = n(moles) / L(volume,Liters)

to work with, but how can I use the given molarity to find molarity of Na₂CO₃?

[Jzalkm]

how about writing the balanced ionic equations showing the precipitation of the two hard ions by formation of their carbonates.

[Lurifax]

Would it be something like this?:

H₂O + Ca²⁺ + Mg²⁺ + 2Na₂CO₃ 

--> H₂O + Ca(CO₃)₂ + Mg(CO₃)₂ + 2Na²⁺

I'm new to this material

[Jzalkm]

just trying to help.
congrats but would have written each one separately.
for example Mg²⁺ + CO₃^2- --> MgCO₃
same for Ca²⁺

careful with the formula though: "Ca(CO3)₂"
another equation would be when Na₂CO₃ gets ionised in solution.

[Lurifax]

ok, so I have:

H20 + Mg²⁺ + NaCO3 ---> H20 + MgCO3 + Na²⁺

H2O + Ca²⁺ + NaCO3 ---> H20 + CaCO3 + Na²⁺

In order to find the mass of NaCO3 I have to find a molarity i think.
Do you know how I would go about finding this?

[Jzalkm]

H20 + Mg²⁺ + NaCO3 ---> H20 + MgCO3 + Na²⁺

water cancels out on both sides.
sodium carbonate being soluble dissociates into its ions. So, the sodium ions will cancel out as in:

for example Mg²⁺_(aq) + CO₃^2- _(aq) --> MgCO3 _(s)

you may do the same for calcium.
afterwards, you may write the equation for sodium carbonate forming its ions in solution.
careful with formula of sodium carbonate.."NaCO₃"

[Borek]

H20 + Mg²⁺ + NaCO3 ---> H20 + MgCO3 + Na²⁺

What is the water for?