# 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, $N {a}_{2} C {O}_{3}$. $N {a}_{2} C {O}_{3} + 2 H C {l}_{2} \to 2 N a C l + C {O}_{2} + {H}_{2} O$

Jan 2, 2018

0.09558 M

#### Explanation:

First,
convert the grams of $N {a}_{2} C {O}_{3}$ to moles

to do so, divide the given grams by the molar mass of $N {a}_{2} C {O}_{3}$

0.1593 g / 106 g/mol = 0.001503 mol $N {a}_{2} C {O}_{3}$

Second,
use molar ratio to find the moles of $H C l$

0.001503 mol $N {a}_{2} C {O}_{3}$ x $\frac{2 m o l H C l}{1 m o l N {a}_{2} C {O}_{3}}$ = 0.003006 mol $H C l$

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

$c = \frac{m o l}{L} = \frac{0.003006 m o l}{0.03145 L}$ = 0.09558 M