Question

Sodium carbonate (Na2CO3) is available in very pure form and can be used to standardize acid solutions. What is the molarity of an HCl solution if 24.3 mL of the solution is required to react with 0.246 g of Na2CO3?

Answer 1

Answer:

Molarity of HCl = 0.19 M

Explanation:

Moles of $Na_2CO_3$:-

Mass = 0.246 g

Molar mass of $Na_2CO_3$ = 105.988 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.246\ g}{105.988\ g/mol}$

$Moles_{Na_2CO_3}= 0.002321\ mol$

According to the given reaction:-

$Na_2CO_3+2HCl\rightarrow 2NaCl+H_2O+CO_2$

1 mole of $Na_2CO_3$ react with 2 moles of HCl

0.002321 mole of $Na_2CO_3$ react with 2*0.002321 moles of HCl

Moles of HCl = 0.004642 moles

Given that volume = 24.3 mL

Also,  

$1\ mL=10^{-3}\ L$

So, Volume = 24.3 / 1000 L = 0.0243 L

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.004642}{0.0243}$

Molarity of HCl = 0.19 M