3 posts • Page 1 of 1
So starting with the current concentration of OH- in the diluted solution you do pH+pOH=14 so pOH=14-13.25=0.75. Then 10^-0.75=[OH-] in the diluted solution. This comes out to be 0.178 M or 0.18 M with sig figs. From there, we have to work backwards to get to the original solution. So we know the concentration we have now arose from diluting to 500 mL. Therefore we multiply 0.18mol/L by 0.5L to give us 0.09 mol. Following this, we know that our original concentration was diluted from 5 mL to 500 mL and thus to get back to our original 5mL solution (and therefore our original concentration) we divide our moles by .005 (.09/.005) and this gives us the initial concentration of 18M.
The easiest way to do this is to use the dilution formula that we used a lot in Chem 14A: since moles stay the same (n1=n2), M1V1=M2V2, where M is molarity and V is volume. Plugging in the numbers, you would get M1(5.00mL)=(0.18M)(500.0mL) --> M1=18M.