Have any of you taken AP Chemistry? I'm staring at this titration problem and I have no idea what...is going on. If you can put it in simple terms, it would be MUCH appreciated.
Q: In the laboratory 7.52 g of Sr(NO3)2 is dissolved in enough water to form 0.750 L. A 0.100-L sample is withdrawn from this stock solution and titrated with a 0.0425 M solution of Na2CrO4. What volume in mL of Na2CrO4 solution is needed to precipitate all the Sr2+(aq) as SrCrO4?
A:
FW(Sr(NO3)2)=211.63 g/mol -> this is the molecular weight of the strontium nitrate
n = 7.52 g/(211.63 g/mol) = 0.0355 mol Sr2+ -> this is dividing the mass of the strontium nitrate by the molecular weight of the nitrate to get the total mols
M = n/V = 0.0355 mol/0.750 L = 0.0474 M -> this is dividing the strontium nitrate by the volume of the initial mix to figure out its molarity
naliquot=MV=(0.0474 M)(0.100 L)=4.74 x 10-3 mol Sr2+=moles of CrO42- needed -> this is taking the molarity of the solution (which stays the same even if you take 100ml from the initial solution) and dividing it by the volume of the extracted portion to figure out how many mols of the strontium nitrate comes through. it is then infered that the same amount of strontium that gets put in will equal the amount of strontium chromate as per the equation Sr2+ + CrO42- -> SrCrO4
V = n/M = 4.74 mmol/0.0425 M = 111 mL ->this takes the number of mols of chromate needed and divides it by the molarity of the sodium chromate solution to get a volume
if you need any more help just ask, im majoring in chemical engineering lol
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