Question

What is the molarity of the hydrochloric acid if 31.00 mL of HCl is required to neutralize 0.525 g of sodium carbonate? 2HCl(aq)+Na2CO3(aq)→2NaCl(aq)+H2O(l)+CO2(g)

Answer 1

Answer: The molarity of HCl solution is 0.08 M.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Given mass of sodium carbonate = 0.525 g

Molar mass of sodium carbonate = 106 g/mol

Putting values in above equation, we get:

`\text{Moles of sodium carbonate}=(0.525g)/(106g/mol)=0.005mol`

The chemical reaction for the formation of chromium oxide follows the equation:

`2HCl(aq.)+Na_2CO_3(aq.)\rightarrow 2NaCl(aq.)+H_2O(l)+CO_2(g)`

By Stoichiometry of the reaction:

1 mole of sodium carbonate reacts with 2 moles of HCl

So, 0.005 moles of sodium carbonate will react with =
`(1)/(2)* 0.005=0.0025mol` of HCl

To calculate the molarity of HCl, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}`

We are given:

Moles of HCl = 0.0025 moles

Volume of solution = 31.00 mL = 0.031 L (Conversion factor: 1L = 1000 mL)

Putting values in above equation, we get:

`\text{Molarity of HCl}=(0.0025mol)/(0.031L)=0.08M`

Hence, the molarity of HCl solution is 0.08 M.