Question

What volume of 0.011 M HCl is required to titrate a solution containing 282 mL of 0.011 M Na2CO3 and 282 mL of 0.011 M HCl?

A) 141 mL
B) 282 mL
C) 423 mL
D) 564 mL

Answer 1

The volume of 0.011 M HCl required to titrate the solution, we can use the balanced chemical equation between HCl and Na2CO3. From the equation, we can see that one mole of HCl reacts with one mole of Na2CO3. Therefore, the volume of 0.011 M HCl required to titrate the solution is 282 mL.

Step-by-step explanation:

To find the volume of 0.011 M HCl required to titrate the solution, we can use the balanced chemical equation between HCl and Na2CO3:

HCl + Na2CO3 -> 2NaCl + H2O + CO2

From the equation, we can see that one mole of HCl reacts with one mole of Na2CO3. Since both HCl and Na2CO3 have the same molarity and volume, we can calculate the moles of Na2CO3 present in the solution as:

Moles of Na2CO3 = Molarity * Volume = 0.011 M * 0.282 L = 0.003102 mol

Since the stoichiometry is 1:1, we need 0.003102 mol of HCl to react with the Na2CO3. To find the volume of 0.011 M HCl required, we can use the formula:

Volume of HCl = Moles of HCl / Molarity = 0.003102 mol / 0.011 M = 0.282 L = 282 mL

Therefore, the volume of 0.011 M HCl required to titrate the solution is 282 mL, which corresponds to option B.