# adding 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 x10^-3 M Na2CO3wpuld produce how many grams of SrCO3 precipitate?ksp of SrCO3 = 1.1 x 10^-10

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

From 50.0 mL of 10-3 M Na2CO3 we will have50*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 theKsp value of 1.1*10-10 so precipation will occur and allthe excess Sr+2 and CO32- willform precipitate such that remaining concentration ofSr+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 presentbefore precipitation – number of moles in solution afterprecipatation = 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

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