Chemistry
posted by Scott on .
A sample of 6.41 grams of NaOH is dissolved into 675 mL of aqueous 0.250 M NaOH (assume no volume change). This solution is then poured into 2.28 gallons of water. (You may assume that the two volumes can be added.) What is the concentration of NaOH in the final solution?
Answer in units of M.

Convert 6.41g NaOH to moles. moles = grams/molar mass. Call this soln 1.
Convert the 0.25O M solution to moles. moles = M x L = ? Call this soln 2.
Convert 2.28 gallons H2O to L.
Add moles soln 1 to moles soln 2. Then total moles NaOH/L (from the 2.28 gallons) = M of the final soln with respect to NaOH. 
Mr. Bob, I got 0.038119 and it's wrong.

Unfortunately you didn't show your work so I can't find the error for you. I expect you converted 2.28 gallons incorrectly (or you may have forgotten to add in the 675 mL to the 2.28 gallons). My answer is 0.03535M which rounds to 0.0354M to three s.f.

Hint: Remember that reactions happen in
moles. So figure out the moles that reacted
then use the molarity formula to continue
solving the problem.
What volume of 0.699 M K3PO4 is required
to react with 68 mL of 0.661 MMgCl2 accord
ing to the equation
2K3PO4 + 3MgCl2 ! Mg3(PO4)2 + 6KCl
Answer in units of mL