adding 50.0 mL of 1.0 x 10^-3 M Sr(NO3)2 and 50.0 mL of 1.0 x 10^-3...

adding 50.0 mL of 1.0 x 10^-3 M Sr(NO3)2 and 50.0 mL of 1.0 x 10^-3 M Na2CO3wpuld produce how many grams of SrCO3 precipitate? ksp of SrCO3 = 1.1 x 10^-10

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Expert Solution

From 50.0 mL of 10-3 M Sr(NO₃)₂ we will have 50*10^-3 millimoles of Sr⁺².

From 50.0 mL of 10-3 M Na₂CO₃ we will have 50*10^-3 millimoles of CO₃^2-.

Total volume after mixing = 100ml

The concentration of Sr⁺² = CO₃^2- = 50*10-3/100 = 0.0005M

[Sr⁺²][CO₃^2-] = 0.0005*0.0005 = 25*10^-8 = 2.5*10^-7 which is greater than the Ksp value of 1.1*10^-10 so precipation will occur and all the excess Sr⁺² and CO₃^2- will form precipitate such that remaining concentration of Sr⁺² and CO₃^2- is such that [Sr⁺²] = [CO₃^2-] and [Sr⁺²][CO₃^2-] = Ksp.

so, lets say after precipitation the new concentration of [Sr⁺²] = [CO₃^2-] = s

so, s² = Ksp = 1.1*10^-10 so, s = 1.048*10^-5M so, number of moles = s*volume in L = 1.048*10^-6

So, the number of moles in precipitate = number of moles present before precipitation - number of moles in solution after precipatation = 50*10^-6 - 1.048*10^-6 = 48.95*10^-6 moles

Weight of 48.95*10^-6 moles of SrCO₃ = 48.95*10^-6*147.63 = 7.22*10^-3 g

queen_honey_blossom answered 1 year ago

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