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In: Chemistry

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

Solutions

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 CO32-.

Total volume after mixing = 100ml

The concentration of Sr+2 = CO32- = 50*10-3/100 = 0.0005M

[Sr+2][CO32-] = 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 CO32- will form precipitate such that remaining concentration of Sr+2 and CO32- is such that [Sr+2] = [CO32-] and [Sr+2][CO32-] = Ksp.

so, lets say after precipitation the new concentration of [Sr+2] = [CO32-] = 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 2 years ago
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