Question

if 54.8 mL of BaCL2 solution is needed to precipitate all the sulfate in 554 mg sample of NA2S04 (forming BAS04) what is the normality of the solution?

Answer 1

To find the normality of the BaCl2 solution, calculate the moles of BaCl2 and the volume of the solution. The normality of the BaCl2 solution is 0.071 N.

Step-by-step explanation:

To find the normality of the BaCl2 solution, we need to calculate the moles of BaCl2 and the volume of the solution.

First, we calculate the moles of BaCl2 using the formula:

Moles of BaCl2 = (mass of BaCl2 / molar mass of BaCl2)

Given that the mass of the sample of Na2SO4 is 554 mg and the molar mass of Na2SO4 is 142.04 g/mol, we have:

Moles of Na2SO4 = (0.554 g / 142.04 g/mol) = 0.003898 mol

Since 1 mole of BaCl2 reacts with 1 mole of Na2SO4 to form BaSO4, the number of moles of BaCl2 is also 0.003898 mol.

Next, we calculate the volume of the BaCl2 solution using the formula:

Volume of BaCl2 solution = (Volume of BaCl2 / Moles of BaCl2)

Given that the volume of the BaCl2 solution is 54.8 mL, we have:

Volume of BaCl2 solution = (54.8 mL / 0.003898 mol) = 14083 mL/mol

To convert the volume to liters, we divide by 1000:

Volume of BaCl2 solution = (14083 mL/mol) * (1 L/1000 mL) = 14.083 L/mol

Finally, we can calculate the normality of the BaCl2 solution using the formula:

Normality = (Number of equivalents / Volume of solution)

Since BaCl2 is a 1:1 electrolyte, it has 1 equivalent of BaCl2 per mole of BaCl2. Therefore, the normality of the BaCl2 solution is:

Normality = (1 equivalent / 14.083 L/mol) = 0.071 N

Answer 2

According to the statement, we have the following balanced reaction:

`BaCl_(2(aq))+Na_2SO_(4(aq))\rightarrow BaSO_(4(s))+2NaCl_((aq))`

We see that the ratio BaCl2 to Na2SO4 is 1/1, that is, the same amount of moles of both reactants are needed to react completely. Now, to find the normality of the solution we will follow the following steps.

1. We find the moles of Na2SO4 by dividing the given mass by its molar mass.

2. We find the moles of BaCl2 by the stoichiometry of the reaction.

3. We apply the normality equation. Normality indicates the amount of solute equivalents in 1 liter of solution, it is very similar to molarity, but here we will do a little conversion.

Let's proceed with the calculation.

1. Moles of Na2SO4

`\begin{gathered} molNa_2SO_4=givengNa_2SO_4*(1molNa_2SO_4)/(MolarMass,gNa_2SO_4) \\ molNa_2SO_4=554mgNa_2SO_4*(1g)/(1000mg)*(1molNa_2SO_4)/(142.04gNa_2SO_4)=3.90*10^(-3)molNa_2SO_4 \end{gathered}`

2. Moles of BaCl2

It will be the same moles of Na2SO4, so moles of BaCl2 will be 3.90x10^-4 mol

3. Normality of the solution

It is known that in 1 mol of BaCl2 there are 2 equivalents. So, we have:

`N=(3.9*10^(-4)molBaCl_2)/(54.8mL)*(1000mL)/(1L)*(2eq)/(1molBaCl_2)=0.142N`

The normality of the solution will be: 0.142N