A 25.0 mL solution of KOH is neutralized with 39.5 mL of 0.100 M HBr. What is the concentration of the original KOH solution?

Answer 1

Hey Sydney!

For this acid-base problem, we ultimately want to determine the number of OH- ions created by 25.0 mL of the KOH solution during the neutralization reaction. We can calculate this since we know that the number of OH- ions in 25.0 mL of the KOH solution is equal to the number of H+ ions in 39.5 mL of the 0.100 M HBr solution.

First, we want to multiply the molarity of the HBr solution by the volume of solution used (in liters, or 0.0395 L). This will give us the moles of HBr in the solution, which is also equivalent to the moles of H+ ions created in the acid-base reaction (since HBr is a strong acid, it dissociates completely).

moles H+ = (0.100 M)(0.0395 L) = 0.00395 moles H+

Next, we want to relate this to the moles of OH- created by the 25.0 mL of KOH solution. Just like before, we multiply the molarity of the solution (which we will represent by 'x' since it is still unknown) by the volume of KOH solution used, or 0.0250 L. This is equal to the moles of H+ ions that we previously calculated.

moles OH- = (x)(0.0250 L) = 0.00395 moles OH-

Finally, we can solve for 'x' by dividing both sides by 0.0250 L. This will give us our final answer.

molarity of KOH solution = x = (0.00395 moles OH-)/(0.0250 L) = 0.158 M

Reported to three sig figs. Hope this helps!

Chris Coats