Respuesta :
Solution :
It is given that dilute hydrochloric acid is titrated with sodium carbonate solution.
[tex]$10 \ cm^3 $[/tex] of a 0.100 [tex]$mol/dm^3$[/tex] of hydrocholric acid is used.
Thus, 10 mL of the 0.1 molar dilute hydrochloric acid is titrated with 16.2 mL of Sodium carbonate solution.
The equation would be :
[tex]$Na_2CO_3 (aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + CO_2(g) +H_2O (l)$[/tex]
Now we know that that the
Molarity = [tex]$\frac{n}{v} \times 100$[/tex]
For hydrochloric acid solution,
[tex]$0.1=\frac{n}{10} \times 100$[/tex]
∴ n = 0.01 moles.
Now for one mole of [tex]$Na_2CO_3$[/tex], two moles of HCl acid is used.
For one mole of HCl, one-half mole of [tex]$Na_2CO_3$[/tex] is required.
Therefore, for 0.01 mole of HCl, we require 0.005 mole of [tex]$Na_2CO_3$[/tex].
Hence, 0.01 mole of hydrochloric acid and 0.005 mole of [tex]$Na_2CO_3$[/tex] solution is used.
The color turns red when methyl orange indicator is used in HCl acid.
