Question

A solution is prepared by adding 0.300 g of solid NaCl to 50.0 mL of 0.100 M CaCl2. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.

Answer 1

The molarity of the chloride ion is 0.3026 M ≈ 0.303 M

Step-by-step explanation:

Step 1: Data given

Mass of solid NaCl = 0.300 grams

Volume of CaCl2 = 50.0 mL

Molarity of CaCl2 solution = 0.100 M

Volume of the final solution = 50.0 mL

Step 2: The balanced equation

NaCl → Na+ + Cl-

CaCl2 → Ca^2+ + 2Cl-

CaCl2 → NaCl2 + CaCl

Step 3: Calculate the number of moles Cl2 in CaCl2

In 1 mol CaCl2 we have 1 mol Ca^2+ and 2 mol Cl-

Moles CaCl2 = molarity ¨volume

Moles CaCl2 = 0.100 M * 0.050 L

Moles CaCl2 = 0.005 moles

In 1 mol CaCl2 we have 1 mol Ca^2+ and 2 mol Cl-

For 0.005 moles CaCl2 we have 2*0.005 = 0.010 moles Cl-

Step 4: Calculate moles Cl in NaCl

Moles NaCl = mass NaCl / molar mass NaCl

Moles NaCl = 0.300 grams / 58.44 g/mol

Moles NaCl = 0.00513 moles

In 1 mol NaCl we have 1 mol Na+ and 1 mol Cl-

In 0.00513moles NaCl we have 0.00513 moles Cl-

Step 5: Calculate the total number of moles Cl-

Total moles Cl- = 0.010 moles + 0.00513 moles

Total moles Cl- = 0.01513 moles Cl-

Step 6: Calculate the molarity of Cl-

Molarity = moles / volume

Molarity Cl- = 0.01513 moles / 0.050 L

Molarity Cl- = 0.3026 M

The molarity of the chloride ion is 0.3026 M ≈ 0.303 M

Answer 2

0.3026 M

Step-by-step explanation:

Whe we have start with the ionization reaction of
`CaCl_2`, so:

`CaCl_2_(_a_q_)~--->~2Ca^+^2~+~Cl^-^1`

The molar ratio between
`CaCl_2` and
`Cl^-^1` is 2:1. With this information we can calculate the concentration of chloride ions in the 0.1 M solution:

`0.1~M~CaCl_2(2)/(1)=~0.2~M~Cl^-^1`

With the concentration of chloride ions and the volume we can calculate the total moles of
`Cl^-^1` (lets keep in mind that 50 mL = 0.05L):

`0.2~M=(#~mol)/(0.05~L)`

`mol~=~0.2*0.05=~0.01~mol~Cl^-^1`

We have so far 0.01 mol of
`Cl^-^1`. With the information of NaCl we can calculate the moles of
`Cl^-^1` present in 0.3 of NaCl (the molar mass of NaCl 58.44 g/mol):

`0.3~g~NaCl(1~mol~NaCl)/(58.44~g~NaCl)=0.00513~mol`

Now we can calculate the total moles of
`Cl^-^1`:

`0.00513~mol~+~0.01~mol=~0.01513~mol~Cl^-^1`

With the total moles and the volume (50mL= 0.05L) we can calculate the concentration:

`M=(0.01513~mol~Cl^-^1)/(0.05~L)=~0.3026~M~of~Cl^-^1~ions`

The total concentration of
`Cl^-^1` ions is 0.3026 M