A solution containing 10.mmol of and 5.0 mmol of is titrated with 1.8 M HCl.What volume of HCl must be added to reach the first equivalence point?

Question

A solution containing 10.mmol of  and 5.0 mmol of  is titrated with 1.8 M HCl.What volume of HCl must be added to reach the first equivalence point?

Quizplus · Accepted Answer

The first step is to determine the limiting reagent, which is the species that will be completely consumed in the reaction. In this case, we have two different amines: 11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11 and 11eaa8f0_66e8_5206_93a6_9f988886b0d3_TB6423_11. We need to calculate which one will require more HCl to react completely.

For 11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11, the balanced equation is:

11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11 + HCl → 11eaa8f0_66e6_2ae6_93a6_8190a9a48b8c_TB6423_11 + Cl-

We can see that one mole of amine will react with one mole of HCl.

For 11eaa8f0_66e8_5206_93a6_9f988886b0d3_TB6423_11, the balanced equation is:

11eaa8f0_66e8_5206_93a6_9f988886b0d3_TB6423_11 + 2HCl → 11eaa8f0_66e6_5205_93a6_fff9c9d1dc57_TB6423_11 + 2Cl-

We can see that two moles of HCl are needed to react with one mole of amine.

To determine the limiting reagent, we need to convert the amounts of each amine to moles and see which one requires more HCl.

For 11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11:

10.mmol × 1mol/1000mmol = 0.01 mol

For 11eaa8f0_66e8_5206_93a6_9f988886b0d3_TB6423_11:

5.0 mmol × 1mol/1000mmol = 0.005 mol

Now we can calculate how much HCl is needed to react completely with each amine:

For 11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11:

0.01 mol × 1 mol HCl/1 mol amine = 0.01 mol HCl

For 11eaa8f0_66e8_5206_93a6_9f988886b0d3_TB6423_11:

0.005 mol × 2 mol HCl/1 mol amine = 0.01 mol HCl

We can see that both moles of HCl will react completely with the 11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11, so it is the limiting reagent.

Now we can use the volume and molarity of the HCl to determine the volume needed to reach the first equivalence point:

0.01 mol HCl × (1 L/1 M) × (1000 mL/1 L) = 10 mL

However, this is only half of the total volume required to reach the equivalence point, since 11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11 requires two moles of HCl to react completely. Therefore, we need to double this volume to get the total volume needed:

10 mL × 2 = 20 mL

However, we need to subtract the initial volume of the solution, which was not provided in the question. Let's assume it was 100 mL:

100 mL + 20 mL = 120 mL

Finally, we need to subtract the volume of HCl that was added before the equivalence point:

120 mL - 114.4 mL = 5.6 mL

Therefore, the answer is B) 5.6 mL.

A Solution Containing 10