Me and my TA have been working this problem, and we're gettingthe same answer fo5r Q but the computer is saying that we arewrong. Can someone help? I'd appreciate it. If 25.0 mL of 0.10 MBa(NO3)2 are added to 41.0 mL of 0.15 MNa2CO3, will BaCO3 precipitate? Find Q: Me and my TA have been working this problem, and we're gettingthe same answer fo5r Q but the computer is saying that we arewrong. Can someone help? I'd appreciate it. Find Q:
Explanation / Answer
Well, first find the new molarity of each molecule: Since 25 mL is added to 41 mL, the new volume is 66 mLor .066 L. Now find how many moles of each molecule you have. (.025)(.1) = .0025 moles ofBa(NO3)2(.041)(.15) = .00615 moles ofNa2CO3 Now molarity of each: .0025/.066 = .038 M Ba(NO3)2.00615/.066 =.093 M Na2CO3 The chemical equation for BaCO3 is BaCO3(s) Ba+(aq) +CO3-(aq) Compute Q and compare it to Ksp (which in this caseis 8.1 x 10-9). A precitate will only form if Q> Ksp. Q = [Ba+][CO3-] = (.038)(.093) = .003534 M Q > Ksp ; precipitate forms