Chemistry: pH after adding water

[theCandyman]

I am trying to find what the new concentration of [tex]OH^{-}[/tex] in a solution of potassium hydroxide (pH = 11.65) is after diluting it with water to six times the original volume. It is at 298.15 K (25 degrees Celcius).

I have found the concentration of [tex]H_{3}O^{+}[/tex] and I guessed that I just multiply the volume times six (divide the concentration by six) to find the new concentration, would this be a correct way to approach this problem?

 

[Sirus]

[tex]c_{1}v_{1}=c_{2}v_{2}[/tex]

where c is concentration and v is volume. Find the initial [OH^-], then use the above formula and the information given to find the new concentration.

 

[thepatientmental]

Yes, your approach is correct. The concentration of OH^{-} in a solution of potassium hydroxide can be calculated by taking the negative logarithm of the hydroxide ion concentration, which is equal to the concentration of potassium hydroxide. So, if you dilute the solution by a factor of six, the concentration of OH^{-} will decrease by a factor of six as well. This means that the new concentration of OH^{-} can be calculated by dividing the original concentration by six. Keep in mind that this calculation assumes ideal behavior and does not take into account any potential changes in temperature or pressure. Overall, your approach is a good way to estimate the new concentration of OH^{-} in the solution after dilution with water.