Assuming you weighed out a 2.3684 g sample of your unknown, and it took 35.63 mL

Question

Assuming you weighed out a 2.3684 g sample of your unknown, and it took 35.63 mL of a 0.1025 M HCI titrant to reach the endpoint, what is the weight percent of Na2CO3 and NaHCO3 in your unknown sample? Solve the simultaneous equations Hints: -Set Na2CO3 = X -Eqn A: bicarbonate carbonate diluted weighted mass (remember that the mass is not 2.3684, you diluted it before you titrated it. Calculate the diluted g of unknown) -Convert the g of Na2CO3 to mol of HCI -Convert the g of NaHCO3 to mol of HCI -mol HCl mol HCl from Na2CO3 + mol HCI from NaHCO3 -Before taking the final weight percent, undo the dilution to get back to the original wt % in the unknown -Na,COg + 2HCI 2 NaCl + H2O + CO2 -NaHCO3 + HCL NaCl + H2O + CO2

Explanation / Answer

answer:

molarity = moles / volume of solution in L

So,

Moles of HCl reacted = 0.1025 * 0.03563 = 3.65*10-3 moles

Assume 'x' g of Na2CO3 is present and 'y' g of NaHCO3 in the 25mL aliquot.

So,

moles of Na2CO3 = x/106

moles of NaHCO3 = y/84

1 mole Na2CO3 requires 2 moles HCl, and 1 mole NaHCO3 requires 1 moles HCl

So,

2*(x/106) + 1*(y/84) = 3.65*10-3

Also, since original volume was 250 mL out of which this aliquot was taken, so actual masses are 10*x and 10*y.

Thus,

10*x + 10*y = 2.3684

Solve the above two equations to get the values of x and y

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.