Hey i need u to fill out the missing data from the data I\'ve found. Also the se

Question

Hey i need u to fill out the missing data from the data I've found. Also the second one I want u to find the standard deviation thanks

Desk No. Date Lab Sec. Name A. A Standard 0.01 M Disodium Ethylenediaminetetraacetate. Naabar, solution Calculate the mass of Na H.Y.2Hzo required to prepare 250 mL ofa001 M Na,HaY solution. Sce Part A 1. 372.24 1000 mL. 0.002 mol 9306 n Trial 3 Trial 2 Trial 25.0 1. Volume of standard Ca solution (mL) 2. Concentration of standard ca solution (molL) 3. Mol ca mol NaoH2Y (mol) 4. Buret reading, inirial (ml) 5. Buret reading, final (mL) 6. volume of NazH.Y titrant (mL) 7. Molar concentration of Na H.Y solution (moura 0 0.22 mol 8. Average molar concentration of Na,HY solution (molyzy B. Analysis of water sample Trial 3 Trial 1 Trial 2 2.5 mL Sm 2.3 m L 1. sample volume (ml) 0 mL 17.0 m 2. Buret reading, initial (mL) 24, m 3. Buret reading, final (mL) 4. volume of NadHiY titrant (mL) Experiment 21 261

Explanation / Answer

B)Analysis of water sample

5)Moles of Na2H2Y=Molarity of Na2H2Y*volume of Na2H2Y used up as titrant=0.225 mol/L*6ml*(1L/1000ml)=0.00135 moles=moles of hardening ions

similarly ,run2,

Moles of Na2H2Y=0.225 mol/L*6ml*(1L/1000ml)=0.00157 moles=moles of hardening ions

run 3,

Moles of Na2H2Y=0.225 mol/L*11ml*(1L/1000ml)=0.00247 moles=moles of hardening ions

6)run 1

molar mass of Ca2+=40g/mol

mass of hardening ions =0.00135 moles*40g/mol=0.054g Ca2+

molar mass of CaCO3=100g/mol

40g/mol Ca2+/100 g/mol CaCO3 =0.054 g Ca2+/x g CaCO3

x=0.135 g

run 3) mass of eq Ca2+ ions=0.00157 moles*40g/mol=0.0628 g/mol

So,

40g/mol Ca2+/100 g/mol CaCO3 =0.0628 g Ca2+/x g CaCO3

x=0.157 g

run 2)

mass of eq Ca2+ ions=0.00247 moles*40g/mol=0.0988 g/mol

So,

40g/mol Ca2+/100 g/mol CaCO3 =0.0988 g Ca2+/x g CaCO3

x=0.247 g

6) run 1) concentration CACO3=0.135g/(25ml+6ml)=4.534 g/L=453.4mg/L=4354.8ppm

run 2)concentration CaCO3=0.157g/(25ml+11ml)=4.534 g/L=453.4mg/L=4361.1ppm

run 3) concentration CaCO3=0.247g/(25ml+7ml)=4.534 g/L=453.4mg/L=7718.7ppm

7)Av ppm CaCO3=4354.8+4361.1+7718.7/3=5478.2 ppm

8) 1gpg or grain per gallon=64.8mg/3.79L=17.1 ppm

so run 1) Av gpg=5478.2/17.1=320.362 gpg

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