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#
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,
1 Answer
Jan 2, 2018
0.09558 M
Explanation:
First,
convert the grams of
to do so, divide the given grams by the molar mass of
0.1593 g / 106 g/mol = 0.001503 mol
Second,
use molar ratio to find the moles of
0.001503 mol
Third,
use the Molarity formula to calculate the concentration of HCl