Question

chemistry homework helpA 100mL sode ash solution was prepared with 5.0g of sodium carbonate/ MW=106 g/mol what is the concentration in M of the solution? What is the concentration of sodium ions?

Answer 1

Answer: the concentration of the prepared solution is 0.47 M and the concentration of Na+ ions in this solution is 0.97 M.

Step-by-step explanation:

The question requires us to calculate the molar concentration of a prepared sodium carbonate (Na2CO3) solution and the concentration of sodium ions (Na+) in this solution.

The following information was provided by the question:

mass of salt used = m = 5.0 g

volume of solution = V = 100 mL

molar mass of sodium carbonate = MM = 106 g/mol

We can calculate the molar concentration of the solution using the following equation:

`M=\frac{n\text{ \lparen mol\rparen}}{V\text{ \lparen L\rparen}}`

where M is the molarity of the solution (in mol/L or M), n is the number of moles (in mol), and V is the volume of solution (in L).

Since the volume was given in mL, we need to convert it to L:

`V=100\text{ mL}*\frac{1\text{ L}}{1000\text{ mL}}=0.1L`

We also need to determine the number of moles of Na2CO3 in 5.0g of this salt:

106 g Na2CO3 ---------------- 1 mol Na2CO3

5.0 g Na2CO3 ----------------- x

Solving for x, we have that there are 0.047 moles of Na2CO3 in 5.0g of this salt.

Now, applying these values to the equation for molar concentration:

`M=(0.047mol)/(0.1L)=0.47M`

Therefore, the concentration of the prepared solution is 0.47 M.

The concentration of Na+ ions can be obtained from the dissociation of Na2CO3:

`Na_2CO_(3(aq))\rightarrow2Na_((aq))^++CO_(3(aq))^(2-)`

According to the dissociation equation, each mol of Na2CO3 produces 2 moles of Na+. Therefore, a 0.47 M Na2CO3 solution should contain 0.47 * 2 = 0.94 M Na+ ions.