Question

During a titration a student found that 20.0cm3 of sodium carbonate solution

neutralised 27.9 cm3 of 0.500 mol/dm3 nitric acid solution.
Calculate the concentration of the sodium carbonate solution in mol/dm3. Give
your answer to 3 significant figures.

Answer 1

Answer: The concentration of
`Na_2CO_3` required is
`0.349mol/dm^3`

Step-by-step explanation:

According to the neutralization law,

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1` = basicity of
`HNO_3` = 1

`M_1` = molarity of
`HNO_3` solution =
`0.500 mol/dm^3`

`V_1` = volume of
`HNO_3` solution =
`27.9cm^3`

`n_2` = acidity of
`Na_2CO_3` = 2

`M_1` = molarity of
`Na_2CO_3` solution =?

`V_1` = volume of
`Na_2CO_3` solution =
`20.0cm^3`

Putting in the values we get:

`1* 0.500* 27.9=2* M_2* 20.0`

`M_2=0.349mol/dm^3`

Therefore, concentration of
`Na_2CO_3` required is
`0.349mol/dm^3`