Answer: The concentration of [tex]Na_2CO_3[/tex] required is [tex]0.349mol/dm^3[/tex]
Explanation:
According to the neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1[/tex] = basicity of [tex]HNO_3[/tex] = 1
[tex]M_1[/tex] = molarity of [tex]HNO_3[/tex] solution = [tex]0.500 mol/dm^3[/tex]
[tex]V_1[/tex] = volume of [tex]HNO_3[/tex] solution = [tex]27.9cm^3[/tex]
[tex]n_2[/tex] = acidity of [tex]Na_2CO_3[/tex] = 2
[tex]M_1[/tex] = molarity of [tex]Na_2CO_3[/tex] solution =?
[tex]V_1[/tex] = volume of [tex]Na_2CO_3[/tex] solution = [tex]20.0cm^3[/tex]
Putting in the values we get:
[tex]1\times 0.500\times 27.9=2\times M_2\times 20.0[/tex]
[tex]M_2=0.349mol/dm^3[/tex]
Therefore, concentration of [tex]Na_2CO_3[/tex] required is [tex]0.349mol/dm^3[/tex]
