Question

Referring to the following information, if a 0.1593 gram sample of sodium carbonate, `Na_2CO_3` requires 31.45 mL of `HCl` solution for complete neutralization, what is the precise concentration of the acid?

A student must prepare a solution of HCl whose concentration will be precisely known. He prepares a liter of solution which is approximately 0.1 M, and uses it to titrate precisely weighed samples of sodium carbonate, `Na_2CO_3`. `Na_2CO_3 + 2HCl_2 -> 2NaCl + CO_2 + H_2O`

Answer 1

0.09558 M

Explanation:

First,
convert the grams of `Na_2CO_3` to moles

to do so, divide the given grams by the molar mass of `Na_2CO_3`

0.1593 g / 106 g/mol = 0.001503 mol `Na_2CO_3`

Second,
use molar ratio to find the moles of `HCl`

0.001503 mol `Na_2CO_3` x `(2 mol HCl)/(1molNa_2CO_3)` = 0.003006 mol `HCl`

Third,
use the Molarity formula to calculate the concentration of HCl

`c = (mol)/L = (0.003006mol)/(0.03145 L)` = 0.09558 M