What is the average molarity of copper in the 100mL dilutions? (MW Cu = 63.542g-

Question

What is the average molarity of copper in the 100mL dilutions? (MW Cu = 63.542g-mol^-1) Complexation Titration of Hard Water 0.9534g EDTA (Purity = 101.6%; H_2O adsorption = 0.7%) is weighed out, dissolved and diluted to 250mL to use in a titration of a water sample to determine water hardness. What is the molarity of EDTA to 4 significant figures? Three 50mL aliquots of an unknown water sample are prepared for titration with the EDTA in question 7. If the volumes of EDTA required for titration are 24.1mL, 25.9mL, 23.3mL, what is the average %(W/W) (assume all Ca) (water hardness)ion in the unknown sample? What is the water hardness in ppm for question 8? Determination of Iron in an Ore 10.) Approximately 1.58g KMnO_4 is dissolved into 500mL deionized H_2O. Three Na_2C_2O_4 samples weighing 0.2112g, 0.2091g, and 0.2015g are titrated with the KMnO_4 with volumes of 31.5mL, 31.0mL, and 30.7mL, respectively. What is the molarity of KMnO_4? (MW Na_2C_2O_4 = 134g middot mol^-1) The KMnO_4 from question 10 is then used to titrate three iron-ore samples, masses of 0.2318g, 0.2153g, and 0.2251g, at volumes of 22.4mL, 20.3mL, and 21.6mL. What is the average % iron (by mass) in the original ore sample? (MW Fe = 54.845g mol middot 1mol^-1.)

Explanation / Answer

(10) mass of Na2C2O4 =(0.2112+0.2091+0.2015)g/3

                                 = 0.2073g

number of Na2C2O4 moles=0.2073g/134gmoi-1

                                        = 1.547 x 10-3 mol

16H+ + 2MnO4- +5C2O42- ----------> 2Mn2+ +8H2O+10CO2

Na2C2O4 : KMnO4=5:2

number of KMnO4 moles= 2/5 x1.547 x 10-3 mol

                                    =6.188 x 10-4 mol

mean volume of KMnO4 spent =(31.5+31.0+30.7)ml/3

                                            31.067ml

molarity of KMnO4 solution= 6.188 x 10-4 mol/ 31.067ml

                                           = 0.0199mlo l-1

(11)mean mass of ore=(0.2318+0.2153+0.2251)g/3

                                =0.2241g

mean volume of KMnO4 spent=(22.4+20.3+21.6)ml/3

                                           =21.43ml

number of KMnO4 spent= 0.0199mlo l-1 x 21.43 x 10-3l

                                  = 4.264 x 10-4mol

16H+ + 2MnO4- +5 Fe----------> 2Mn2+ +8H2O+5Fe2+

KMnO4 : Fe =2:5

number of Fe moles =5/2 x4.264 x 10-4mol

                            =1.066 x10-3mol

mass of ore in the sample=1.066 x10-3mol x 54.845gmol-1

                                    = 58.464 x 10-3g

percentage of ore=( 58.464 x 10-3g/0.2241g) x 100%

                        = 2.608%

                        =

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