Question

Calculate the mass of CaCO₃ required to prepare 500 mL of a standard solution that is 100 ppm in Ca²⁺.
(Include units and use an appropriate number of sig figs.)

Calculate the mass of NaCl required to prepare 500 mL of a standard solution that is 100 ppm in Na⁺.
(Include units and use an appropriate number of sig figs.)

Answer 1

We know that 1 ppm = 1 mg/L.

a) 100 ppm Ca²⁺ = 100 mg/L Ca²⁺.

Atomic mass of Ca = 40.078 g/mol.

Mole(s) of Ca²⁺ corresponding to 100 mg/L = (100 mg/L)*(1 g/1000 mg)*(1 mole/40.078 g) = 0.002495 mole/L Ca²⁺.

We have 500 mL solution; hence, the mole(s) of Ca²⁺ = (0.002495 mol/L)*(500 mL)*(1 L/1000 mL) = 0.0012475 mole.

Consider the ionization of CaCO₃ as below.

CaCO₃ ------> Ca²⁺ + CO₃^2-

As per the stoichiometric equation,

1 mole CaCO₃ = 1 mole Ca²⁺.

Therefore, 0.0012475 mole Ca²⁺ = 0.0012475 mole CaCO₃.

Molar mass of CaCO₃ = (1*40.078 + 1*12.01 + 3*15.9994) g/mol = 100.0862 g/mol.

Mass of CaCO₃ required = (0.0012475 mole)*(100.0862 g/mol) = 0.124857 g ≈ 0.125 g (ans).

b) 100 ppm Na⁺ = 100 mg/L Na²⁺.

Atomic mass of Na = 22.9897 g/mol.

Mole(s) of Na⁺ corresponding to 100 mg/L = (100 mg/L)*(1 g/1000 mg)*(1 mole/22.9897 g) = 0.0043498 mole/L Na⁺.

We have 500 mL solution; hence, the mole(s) of Na⁺ = (0.0043498 mol/L)*(500 mL)*(1 L/1000 mL) = 0.0021749 mole.

Consider the ionization of NaCl as below.

NaCl ------> Na⁺ + Cl^-

As per the stoichiometric equation,

1 mole NaCl = 1 mole Na⁺.

Therefore, 0.0021749 mole Na⁺ = 0.0021749 mole NaCl.

Molar mass of NaCl = (1*22.9897 + 1*35.453) g/mol = 58.4427 g/mol.

Mass of NaCl required = (0.0021749 mole)*(58.4427 g/mol) = 0.1255 g ≈ 0.125 g (ans).