Question

Indicates a concentration of each ion present in a solution formed by mixing the following. 3.40g of NaCl in 43.3 mL of 0.556 M CaCl2 solutionNa^+Ca^2+Cl^-

Answer 1

1) Concentration NaCl

1.1- List the known quantities.

3.40 g NaCl

Volume: 43.3 mL

1.2- Molarity of NaCl

The molar mass of NaCl is 58.4428 g/mol.

`mol\text{ }NaCl=3.40\text{ }g\text{ }NaCl*\frac{1\text{ }mol\text{ }NaCl}{58.4428\text{ }g\text{ }NaCl}=0.0582\text{ }mol\text{ }NaCl`

Convert mL to L

1 L = 1000 mL

`L=43.3\text{ }mL*\frac{1\text{ }L}{1000\text{ }mL}=0.0433\text{ }L`

1.3- Set the equation

`M=\frac{moles\text{ }of\text{ }solute}{liters\text{ }of\text{ }solution}`

1.4- Plug in the known quantities

`M=\frac{0.0582\text{ }mol\text{ }NaCl}{0.0433\text{ }L}=1.34\text{ }M\text{ }NaCl`

The molar concentration of NaCl is 1.34 M NaCl.

2) Concentration of Na

The ion-molecule ratio is 1 mol Na = 1 mol NaCl.

`M_(Na^+)=1.34\text{ }NaCl\text{ }\frac{mol\text{ }NaCl}{L}*\frac{1\text{ }mol\text{ }Na}{1\text{ }mol\text{ }NaCl}=1.34\text{ }*\frac{mol\text{ }Na^+}{L}`

The concentration of Na+ is 1.34 M Na+.

3) Concentration of CaCl2

The ion-molecule ratio is 1 mol Ca (2+): 1 mol CaCl2.

Molarity: 0.556 M CaCl2

Volume: 0.0433 L

`M_(Ca^(2+))=0.556\text{ }\frac{mol\text{ }CaCl_2}{L}*\frac{1\text{ }mol\text{ }Ca^(2+)}{1\text{ }mol\text{ }CaCl_2}=0.556*\frac{mol\text{ }Ca^(2+)}{L}`

The concentration of Ca2+ is 0.556 M Ca2+.

4) Concentration of Cl-

4.1- Moles of Cl- from NaCl

The ion-molecule ratio is 1 mol Cl = 1 mol NaCl.

Volume: 0.0433 L

The molar mass of NaCl is 58.4428 g/mol.

`mol\text{ }NaCl=3.40\text{ }g\text{ }NaCl*\frac{1\text{ }mol\text{ }NaCl}{58.4428\text{ }g\text{ }NaCl}=0.0582\text{ }mol\text{ }NaCl`

We have 0.0582 mol NaCl.

`mol\text{ Cl}^-=0.0582\text{ }mol\text{ }NaCl*\frac{1\text{ }mol\text{ }Cl^-}{1\text{ }mol\text{ }NaCl}=0.0582\text{ }mol\text{ }Cl^-`

We have 0.0582 mol Na from NaCl.

4.2- Moles of Cl- from CaCl2.

Molarity: 0.556 M CaCl2

Volume: 0.0433 L

`M_(Ca^(2+))=0.556\text{ }\frac{mol\text{ }CaCl_2}{L}*\frac{0.0433\text{ }L}{}=0.0241mol\text{ }CaCl_2`

Moles of Cl-

0.0241 mol CaCl2.

The ion-molecule ratio is 2 mol Cl (-): 1 mol CaCl2.

`mol\text{ }Cl^-=0.0241\text{ }mol\text{ }CaCl_2*\frac{2\text{ }mol\text{ Cl}^-}{1\text{ }mol\text{ }CaCl_2}=0.0482\text{ }mol\text{ }Cl^-`

We have 0.0482 mol Cl- from CaCl2.

4.3- Add up moles of Cl- from NaCl and moles of Cl- from CaCl2.

We have 0.0582 mol Na from NaCl.

We have 0.0482 mol Cl- from CaCl2.

0.0582 + 0.0482 = 0.1064 mol Cl-

4.4- Molarity of Cl-

Moles of Cl-: 0.1064 mol

Volume: 0.0433 L

`M=\frac{0.1064\text{ }mol\text{ }Cl^-}{0.0433\text{ }L}=2.46\text{ }M`

The concentration of Cl- is 2.46 M Cl-.

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