Question

2.65 g of sodium carbonate was dissolved in water and made up to 500 mL in a standard flask. 20 mL portions of this solution required 18.50 mL of a solution of hydrochloric acid when titrated using methyl orange indicator.Calculate: (i) The concentration of the standard sodium carbonate solution in mol/L. 

Answer 1

Answer:

0.0500 mol/L.

Explanation:

What is given?

Mass of sodium carbonate (Na2CO3) = 2.65 g,

Molar mass of Na2CO3 = 106 g/mol,

1 L = 1000 mL.

Step-by-step solution:

This problem is asking us to find the molarity of the sodium carbonate solution. Remember that the formula of molarity is the following:

$Molarity\text{ \lparen}M)=\frac{moles\text{ of solute}}{liter\text{s of solution}}=\frac{mol}{L}.$

So we have to find the number of moles of Na2CO3 using its molar mass, like this:

$2.6\text{5 g Na}_2CO_3\cdot\frac{1\text{ mol Na}_2CO_3}{106\text{ g Na}_2CO_3}=0.025\text{ moles Na}_2CO_3.$

So we can replace the data that we have in the formula, but remember that as 1 L equals 1000 mL, 500 mL is the same that 0.5 L:

$Molarity(M)=\frac{0.025\text{ moles}}{0.5\text{ L}}=0.05\text{ mol/L.}$

The answer is that the concentration of sodium carbonate solution is 0.0500 mol/L.