What is the molarity of a KOH solution if 25.0 ml neutralizes 35.0 ml of a 0.200 M HCI solution?

in order to neutralize an acidic solution, there must be an equimolar amount of base (KOH). set up the problem like this:

0.350L HCl (0.200M/L) = x moles of HCl

x moles of HCl = x moles of KOH

x moles of KOH (1/0.25L KOH) = molarity of KOH
0.200M = moles of KOH/.025 = 0.005mol
M = 0.005mol/.035L = 0.143M
At neutralization, the moles of acid equals the moles of base. Moles is equal to the molarity times the volume. Therefore,
(molarity KOH)(volume KOH)=(molarity HCl)(volume HCl)

After multiplying the left side and dividing it by 25.0, you should get .280. The molarity of the KOH is .280M.

The answers post by the user, for information only, FunQA.com does not guarantee the right.

More Questions and Answers:
  • Electrical conductivity & dielectric constant question!!?
  • Which of the following salts when added to Pure H2O will cause a significant change in the PH of the solution?
  • Help plz in definte composition?
  • What is the relationship between the effectiveness of the anatacid and the active ingredients?
  • How is milk of magnesia prepared from magnesium carbonate?
  • Where was potassium discovered?
  • Benzophenone oxime mechanism?
  • Explosive formula?
  • Which process requires energy- passive or active transport?