Question

52. A 0.500 g sample containing Na2CO3, plus inert matter is analyzed by adding 50.0mL of

0.100 M HOL, a slight excess, boiling to remove CO2,and then back-titrating the excess acid
with 0.100 M NaOH. If 5.6mL NaOH is required for the back-titration, what is the percent
Na2CO3 in the sample?

53. A hydogen i solution is analyzed by adding a slight excess of standard KMnO4, solution
a back titrating the unreacted KMnO4 with standard Fe2+ solution. A 0.587g sample of the
H2O2 solution is taken. 25.0mL of 0.0215 M KMnO4, is added, and the back-titration requires
5.10 mL of 0.112 M Fe2+ solution. What is the percent H202 in the sample?

Answer 1

To determine the percent Na2CO3 in the sample, first calculate the moles of Na2CO3 using the balanced equation. Then, calculate the mass of Na2CO3 in the sample. Finally, calculate the percent Na2CO3 in the sample by dividing the mass of Na2CO3 by the total mass of the sample and multiplying by 100%.

Step-by-step explanation:

To find the percent Na2CO3 in the sample, we need to determine the amount of Na2CO3 present in the sample and compare it to the total mass of the sample.

First, we calculate the moles of Na2CO3 present in the sample by using the balanced equation:

C2O4H2(aq) + 2NaOH(aq) → Na2C2O4(aq) + 2H2O(l)

From the back-titration, we know that 0.100 M NaOH reacts with 0.100 M HCl. Given that 5.6 mL of NaOH was required for the back-titration, we can calculate the moles of NaOH used:

Moles of NaOH = (0.100 M) * (0.0056 L) = 0.00056 moles

Since 2 moles of NaOH react with 1 mole of Na2CO3, the moles of Na2CO3 in the sample is:

Moles of Na2CO3 = 0.00056 / 2 = 0.00028 moles

Now we can calculate the mass of Na2CO3 in the sample:

Mass of Na2CO3 = (0.00028 moles) * (105.99 g/mol) = 0.02947 g

Finally, we can calculate the percent Na2CO3 in the sample:

Percent Na2CO3 = (0.02947 g / 0.500 g) * 100% = 5.89%