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



Answer:
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)
(X)(25.0mL)=(.200M)(35.0mL)

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

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