A solution of approximately 0.1 M HCl was standardized against dried, primary st

Question

A solution of approximately 0.1 M HCl was standardized against dried, primary standard sodium carbonate to a omocresol endpoint with boiling. Four determinations were run using pure Na_2CO_3 (FW = 106.0) samples with an eraged mass of 0.2155 g, requiring 40.17 mL. The resulting HC1 solution was then used to titrate four unknown samples with an averaged mass of 0.4622 g, requiring 30.08 mL. Determine the normality of the HCI solution; Calculate the percentage of sodium carbonate in the unknown sample; and Calculate the absolute percent error if the actual percentage is 34.80%. A sample of pure CaCO_3 weighing 0.2500 g is dissolved in hydrochloric acid and the solution diluted to 250.0 ml in volumetric flask. A 25.00 ml aliquot requires 39.11 ml of an EDTA solution for titration. Calculate the Titer of the EDTA solution in terms of mg CaCO_3 per ml of EDTA. A 250.0 ml sample of lake water containing Ca^+2 is titrated with 11.00 ml of the above EDTA solution. Cal hardness of the water in mg of CaCO_3 per ml of lake water. Determine the normality of a solution of KMnO_4 (FW = 158.034) if 7.9935 g of primary standard K_2Cr_2O = 294.202) was used to prepare 0.5000 L of this solution 6e^- + 2Cr^+6 = 2Cr^+3 Calculate the percent iron (At. Wt. = 55.847 g/mole) in an unknown if a 0.9887 g sample required 30.05 ml of the above standard KMnO_4 solution to titrate the iron as Fe^+2 in the sample. 5e^- + MnO_4^-1 + 8H^+ = Mn^+2 + 4H_2O

Explanation / Answer

1) a) Moles of Na2CO3 in the standrd solution=mass/formula mass=0.2155g/106g/mol=0.00203 moles

molarity of Na2CO3=moles of Na2CO3/volume required=0.00203moles/40.17ml=0.00203mol/0.04017L=0.0506mol/L

molarity of HCl=molarity of Na2CO3/basicity=0.0506 mol/L/2=0.0253 M

molarity of HCl=Normality of HCl=0.0253M

b)moles of HCl required to titrate unknown Na2CO3=molarity of HCl*volume of HCl=0.0253M*30.08ml=0.0253 mol/L*0.03008L=7.612*10^-4 moles

moles of Na2CO3 titrated=1/2*moles of HCl=1/2*(7.612*10^-4 moles)=3.806*10^-4 mol

So mass of Na2CO3 titrated=moles of Na2CO3 titrated*formula mass=3.806*10^-4 mol*106g/mol=0.0403g

% of Na2CO3 in the unknown sample=mass of Na2CO3/mass of sample*100=0.0403g/0.4622g *100=8.728%

c)absolute % error=difference in value/actual value *100=(34.8-8.728/34.8)*100=74.9%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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