titration of 25ml of Na2CO3 with 0.352mol/l HCl of 15.2ml what is the molar concetraion of sodium carbonate
To determine the molar concentration of sodium carbonate (Na2CO3), we can use the concept of stoichiometry in a titration reaction. The balanced equation for the reaction between sodium carbonate (Na2CO3) and hydrochloric acid (HCl) is:
2Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O
From the equation, we can see that it takes 2 moles of HCl to react with 1 mole of Na2CO3.
Given information:
Volume of HCl solution (V1) = 15.2 mL = 0.0152 L
Molar concentration of HCl (C1) = 0.352 mol/L
Volume of Na2CO3 solution(V2) = 25 mL = 0.025 L
Now, let's calculate the moles of HCl involved in the reaction:
Moles of HCl (n1) = C1 * V1
= 0.352 mol/L * 0.0152 L
= 0.00534 mol
Since the stoichiometric ratio of HCl to Na2CO3 is 2:1, the moles of Na2CO3 (n2) would be half of the moles of HCl:
Moles of Na2CO3 (n2) = 0.00534 mol / 2
= 0.00267 mol
Finally, we can calculate the molar concentration of Na2CO3 (C2):
Molar concentration of Na2CO3 (C2) = moles of Na2CO3 (n2) / volume of Na2CO3 solution (V2)
= 0.00267 mol / 0.025 L
= 0.1068 mol/L
Therefore, the molar concentration of sodium carbonate (Na2CO3) is approximately 0.1068 mol/L.
To find the molar concentration of sodium carbonate (Na2CO3), we can use the concept of stoichiometry and the equation for the balanced chemical reaction that occurs during the titration.
The balanced equation for the reaction between Na2CO3 and HCl is:
Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O
From the balanced equation, we can see that one mole of Na2CO3 reacts with two moles of HCl.
First, let's determine the number of moles of HCl used in the titration.
Given:
Volume of HCl used: 15.2 mL
Concentration of HCl: 0.352 mol/L
Using the formula: moles = concentration x volume (in liters), we can calculate the moles of HCl used:
moles of HCl = 0.352 mol/L x (15.2 mL / 1000 mL/ L)
= 0.352 mol/L x 0.0152 L
= 0.00534 mol
Since the stoichiometric ratio between Na2CO3 and HCl is 1:2, the number of moles of Na2CO3 used would be half of the moles of HCl:
moles of Na2CO3 = 0.00534 mol / 2
= 0.00267 mol
Now, let's calculate the molar concentration of Na2CO3.
Given:
Volume of Na2CO3 used: 25 mL
Molar concentration = ?
Using the formula: concentration = moles / volume (in liters), we can calculate the molar concentration of Na2CO3:
molar concentration of Na2CO3 = 0.00267 mol / (25 mL / 1000 mL/L)
= 0.00267 mol / 0.025 L
= 0.1068 mol/L
Therefore, the molar concentration of sodium carbonate (Na2CO3) is approximately 0.1068 mol/L.
Na2CO3 + 2HCl ==> 2NaCl + H2O + H2O
mol HCl = M x L = ?
Convert mol HCl to mol Na2CO3 using the coefficients in the balanced equation.
?mol HCl x (1 mol Na2CO3/2 mol HCl) = ?mol HCl x 1/2 = xx mol Na2CO3.
M Na2CO3 = mol Na2CO3/L Na2CO3.