Question

A solid powder is known to be a mixture of NaCl and Na2CO3, but the relative amounts of each compound in the sample are unknown. Sodium carbonate reacts with hydrochloric acid according to the equation: Na2CO3(aq) +2 HCl(aq) ? 2NaCl(aq) + H2O(l)+CO2(g) A solution of the mixture is prepared by adding 10.0g of the mixture to enough water to make 1.0L of solution. It is observed that the above reaction goes to completion (i.e. the solution is neutralized) after the addition of 83.15mL of 0.1174M hydrochloric acid to the 1.0L sample of the solution. What is the concentration of sodium carbonate in the solution before the addition of HCl?

Answer 1

Answer:

Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.

Explanation:

$Na_2CO_3(aq) +2 HCl(aq)\rightarrow 2NaCl(aq) + H_2O(l)+CO_2(g)$

Molarity of HCl solution = 0.1174 M

Volume of HCl solution = 83.15 mL = 0.08315 L

Moles of HCl = n

$molarity=\frac{moles}{Volume (L)}$

$0.1174 M=\frac{n}{0.08315 L}$

$n=0.1174 M\times 0.08315 L=0.009762 mol$

According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.

Then 0.009762 mol of HCl will recat with:

$\frac{1}{2}\times 0.009762 mol=0.004881 mol$

Moles of Sodium carbonate = 0.004881 mol

Volume of the sodium carbonate containing solution taken = 1L

Concentration of sodium carbonate in the solution before the addition of HCl:

$[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L$