EDTA Stanardization The EDTA solution was initially standardized against Mg2+. 2

Question

EDTA Stanardization

The EDTA solution was initially standardized against Mg2+. 25.00mL aliquots of Mg2+ solution were used. A stock Mg2+ solution was created by dissolving 0.4673 g of magnesium nitrate hexahydrate in 250.00mL of distilled water. This solution was diluted by taking 2.5 mL and diluting to 250 mL to create the stock solution used in the titrations given.

This is for an analytical chemistry class. Please show all work and excel sheets. Will rate

EDTA Stand. trail V EDTA (mL) 1 33.07 2 31.89 3 36.51 4 32.18 5 33.21 6 32.26 Problem 2 EDTA Titration Lab Data (80 Points) The following data was used in a series for quantifying several common contaminants in drinking water: Pb2+, 2+ Cu and SO In order to analyze all of these, multiple EDTA titrations were performed. Each will be described individually below. Assume any of this data could be outliers (it is raw, laboratory data), and ALL relevant statistics should be performed! Your task will be to determine if these values (statistically) exceed the levels set by the EPA as safe for a natural body of water. Safe levels are given below (taken from: www.epa.gov): Safe Levels Cu 1.3 ppm Pb 0.015 ppm Hg 0.002 ppm SO4 250 ppm pm" is parts per million and is defined as ug/mL.

Explanation / Answer

1)EDTA standardization-

stock solution

# of moles of Mg(NO3)2-6H2O=mass/molar mass=0.4673g/256.41g/mol=0.00182 moles

molarity of stock=Ms=0.00182moles/0.250L=0.00729M

molarity of diluted Mg2+ =M1

volume of stock=Vs=2.5ml

volume of dil sol=V1=250ml

Ms*Vs=M1*V1

M1=Ms*Vs/V1=7.289*10^-5 M

standardization using Mg2+=0.025L*(7.289*10^-5 M)=1.822*10^-5 moles

Volume of EDTA used up=average=33.19ml

Mg2+ +H2Y^2-- ---->Mg(EDTA)^2- +2H+

Molarity of EDTA=M(EDTA)

M(EDTA)* V(EDTA)=M(Mg2+)*V(Mg2+)

M(EDTA)=M(Mg2+)*V(Mg2+)/V(EDTA)=(7.289*10^-5 M)*25ml/33.19ml=5.49*10^-5M

M(EDTA) standardized=5.49*10^-5M

2) titration of Cu2+

Cu2+ +H2Y^2-- ---->Cu(EDTA)^2- +2H+

M(Cu2+)=M(EDTA)std *V(EDTA)/V(Cu2+)

V(Cu2+)=25ml

V(EDTA)average=16.90ml=

M(Cu2+)=(5.49*10^-5M)*16.9ml/25ml=3.71*10^-5 mol/L

So molarity of Cu2+=3.71*10^-5 mol/L [As Cu and EDTA are in molar ratio 1:1 in the complex]

grams of Cu2+=moles*molar mass=(3.71*10^-5 mol/L)*63.55g/mol=0.0235g/L=2.358 mg/L=2.358ppm(above safe level)

3)Similarly,M(Pb2+)=M(EDTA)std *V(EDTA)/V(Pb2+)=(5.49*10^-5M)*19.26ml/25ml=4.23*10^-5 mol/L

molarityof Pb2+=4.23*10^-5 mol/L =4.23*10^-5 mol/L *207.2g/mol=0.00876 g/L=8.764mg/L=8.764 ppm(above safe level)

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

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