# Assignment 3_Answers - 0.00296 M 0.7435 L 0.00398 = ×...

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Quantitative Analysis Assignment #3 Due 10/15/04 Equilibria and Titrations 1) What mass of nickel sulfide (NiS; FW=90.759) can be dissolved in 5 L of water if the K sp is 1.3x10 -25 ? [ ] [ ] - + - = × = 2 2 25 S Ni 10 3 . 1 sp K if x =moles of nickel dissolved, then 2 25 5 10 3 . 1 = × - x mol 10 8 . 1 12 - × = x Mass of Ni= pg) 160 or ng 0.16 or ( g 10 1.6 g/mol 90.759 mol 10 1.8 -10 12 × = × × - 2) A mixture of sodium oxalate (Na 2 C 2 O 4 ; FW=101.9997) and sodium monoxalate (NaHC 2 O 4 ; FW=80.0179) weighed 1.578 g and was titrated to excess with 48.01 mL of 0.522 M HCl. The excess HCl was titrated with 0.7435 M NaOH and needed 3.98 mL of NaOH to be neutralized. What was the weight percent of each of the two compounds in the original sample? Total moles of HCl used= HCl mols 0.0256 0.522M L 0.04801 = × Moles of excess HCl=mols of NaOH used= excess
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Unformatted text preview: 0.00296 M 0.7435 L 0.00398 = × Moles of HCl used for titration= mols 0.0221 0.00296-0.0256 = Reactions: 4 2 2 4 2 2 O C H 2NaCl 2HCl O C Na + ⇒ + 4 2 2 4 2 O C H NaCl HCl O NaHC + ⇒ + If x =mass of Na 2 C 2 O 4 , then the mass of NaHC 2 O 4 =1.578-x The total moles of HCl used for the titration must equal the total moles of oxalate and monoxalate, so: moles 0.0221 80.0179-1.578 9997 . 101 2 = + x x Cross Multiply with 101.9997&80.0179 3758 . 80 1 101.9997x-160.955 0358 . 160 = + x 19.42 036 . 58 = x 3346 . = x For weight percent: % 2 . 21 100 578 . 1 3346 . O C %Na 4 2 2 = × = % 8 . 78 100 578 . 1 3346 . 578 . 1 O %NaHC 4 2 = ×-=...
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