Strontium sulfate is an insoluble solid. Determine the molar concentration of su

Question

Strontium sulfate is an insoluble solid. Determine the molar concentration of sulfate ions in a solution made by mixing 183 mL of a 1.95 mol/L solution of Strontium nitrate with 225 mL of a 1.84 mol/L solution of Lithium sulfate

Find the mass of solid aAluminum hydroxide that will form when 125 mL of 2.34 mol/L Aluminum acetate is mixed with 215 mL of 1.86 mol/L Sodium hydroxide.

83.4 g of potassium iodide is placed into enough water to make 425 mL of solution . 25.0 mL of this solution is then diluted to a final volume of 125.0 mL. What is the concentration of the final solution? Please and thank you soooo much!

Explanation / Answer

Sr(NO3)2 + Li2SO4 ---------------> SrSO4  ? + 2LiNO3

Stronium nitrate lithium sulphate stronium sulphate lithium nitrate

(insoluble solid- precipitate)

Molarity of Sr(NO3)2 M1 = 1.95 M or mol/lt

Volume of Sr(NO3)2 V1 = 183 ml = 0.183 lts

Molarity of Sr+2 after mixing M2 = M

Volume of Sr+2 after mixing V2 = total volume of solution after mixing = 183 ml +225 ml = 408 ml = 0.408 lts

M1V1 = M2V2

1.95 x 0.183 lts = M x 0.408 lts

M = 0.875

Molarity of Sr+2 after mixing M2 = 0.875 M

The source of SO4-2 ions in the final solution is through Li(SO4)2

Molarity of Li2SO4 M1 = 1.84 M or mol/lt

Volume of Li2SO4 V1 = 225 ml = 0.225 lts

Molarity of SO4-2 after mixing M2 = M

Volume of SO4-2 after mixing V2 = total volume of solution after mixing = 183 ml +225 ml = 408 ml = 0.408 lts

M1V1 = M2V2

1.84 x 0.225 lts = M x 0.408 lts

M = 1.015 M

Molarity of SO4-2 after mixing M2 =1.015 M

2) Al(CH3COO)3 + 3NaOH --------> Al(OH)3 + 3NaCH3COO

Molarity of sodium hyrdroxide = 1.86 mol/l or M

Volume of NaOH = 215 ml = 0.215 lts

So number of moles of NaOH = 1.86 M X 0.215 lts = 0.3999 moles (Molarity = no of moles per lt of volume of solution)

Molarity of aluminium acetate = 2.34 mol/lt or M

volume of aluminium acetate = 125 ml = 0.125 lts

So number of moles of aluminium acetate = 2.34 mol/lt x 0.125 = 0.2925

Molar ratio of aluminium acetate and NaOH is 1:3 (can be observed from the above reaction)

i.e 1 mole of aluminium acetate require 3 moles of NaOH

So 0.2925 moles of aluminium acetate require 3 x 0.2925 moles = 0.8775 moles of NaOH

but the actual moles of NaOH = 0.3999 moles

This shows that NaOH moles are very less in number than aluminium acetate, so the limiting reagent is NaOH  

Molar ratio of NaOH and Al(OH)3 is 3:1

i.e 3 moles of NaOH produce 1 mole of Al(OH)3

So 0.2325 moles of NaOH produce 0.3999/3 = 0.1333 moles of Al(OH)3

So theortical moles of Al(OH)3 = 0.1333 moles (Moles = weight /molar mass)

Molar mass of Al(OH)3 = 78.0036 gm/mol

Mass of Al(OH)3 produced = No. of moles x molar mass = 0.1333 moles x 78.0036 gm/mol =

10.4 gms

c) Weight of KI , m= 83.4 gms

Molar mass of KI , MW = 166.0028 gm/mol

Vol of water to make the solution, V = 425 ml = 0.425 lts

Initial concentration of potassium iodide (KI) is Molarity = (weight/mol wt) / vol of solution in lts

= (83.4 g/166.0028 g/mol ) / 0.425 lts = 1.182 M

Initial molarity of KI, M1 = 1.182 M

Volume of initial molarity of KI solution, V1 = 25 ml = 0.025 lts

Final volume of KI solution, V2 = 125 ml = 0.125 lts

Final concentration of KI M2 = M

M1V1 = M2V2

1.182 M X 0.025 lts = M x 0.125 lts

Final concentration of KI, M = 0.236 M

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