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#

1 Answer
Jan 2, 2018

Answer:

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