Hard water often contains dissolved ca2+ and mg2+ ions. one way to soften water is to add phosphates. the phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from solution. suppose that a solution is 5.5×10−2m in calcium chloride and 9.5×10−2m in magnesium nitrate.
what mass of sodium phosphate must be added to 2.0l of this solution to completely eliminate the hard water ions? assume complete reaction.
would anyone be able to explain to me how to approach this problem? not quite sure where to start. i just know they are asking for the mass of na3po4.

Answer is: mass of sodium phosphate must be added is 32.79 grams.

Balanced chemical reaction 1:

3CaCl₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₃(s) + 6NaCl(aq).

Balanced chemical reaction 2:

3Mg(NO₃)₂(aq) + 2Na₃PO₄(aq) → Mg₃(PO₄)₃(s) + 6NaNO₃(aq).

c(CaCl₂) = 0.055 M; molarity of calcium chloride.

V(CaCl₂) = 2 L; volume of calcium chloride solution.

n(CaCl₂) = c(CaCl₂) · V(CaCl₂).

n(CaCl₂) = 0.055 mol/L · 2 L.

n(CaCl₂) = 0.11 mol; amount of calcium chloride.

From balanced chemical reaction 1: n(CaCl₂) : n(Na₃PO₄) = 3 : 2.

n₁(Na₃PO₄) = 2 · 0.11 mol ÷ 3.

n₁(Na₃PO₄) = 0.073 mol; amount of sodium phosphate in reaction 1.

2) c(Mg(NO₃)₂) = 0.095 M; amount of magnesium nitrate.

V(Mg(NO₃)₂) = 2 L; volume of magnesium nitrate solution.

n(Mg(NO₃)₂) = c(Mg(NO₃)₂) · V(Mg(NO₃)₂).

n(Mg(NO₃)₂) = 0.095 mol/L · 2 L.

n(Mg(NO₃)₂) = 0.19 mol; amount of magnesium nitrate.

From balanced chemical reaction 2: n(Mg(NO₃)₂) : n(Na₃PO₄) = 3 : 2.

n₂(Na₃PO₄) = 2 · 0.19 mol ÷ 3.

n₂(Na₃PO₄) = 0.126mol; amount of sodium phosphate in reaction 2.

3) n(Na₃PO₄) = n₁(Na₃PO₄) + n₂(Na₃PO₄).

n(Na₃PO₄) = 0.073 mol + 0.126 mol.

n(Na₃PO₄) = 0.2 mol; amount of sodium phosphate.

M(Na₃PO₄) = 3Ar(Na) + Ar(P) + 4Ar(O) · g/mol.

m(Na₃PO₄) = n(Na₃PO₄) · M(Na₃PO₄).

m(Na₃PO₄) = 0.2 mol · 163.94 g/mol.

n(Na₃PO₄) = 32.788 g; mass of sodium phosphate.

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.