Hello here is a question that is a bit confusing I hope you can help me solve it and understand it:

Determine the hydronium and hydroxide ion concentrations in a solution that is $10^{-4} M \ce{Ca(OH)2} \text{ AND } 10^{-4} M \ce{HCl}$

Note: the "and" is actually capitalised in my book..

So what I understood from this question is that they are both in one solution. Now I thought of two ways to solve it: A) the concentrations are equal and they are both strong which means they we're neutralise each other completely and end up being a neutral solution and both $\ce{H3O+}$ and $\ce{OH-}$ will be $10^{-7}$.

B) the calcium base has 2 $\ce{OH-}$ while HCl has 1 H+ so the base concentration is double that of the acid which means that the solution is going to be basic but I don't know how can I solve this to find the actual concentrations.

Which way is right and when does an acid and base really react to become neutral?


1 Answer 1


Your second (B) way of thinking is correct.

In the solution there are twice as many OH- ions as H+ ions.

All the H+ from HCL will react with half the OH- from calcium hydroxide.

Half the OH- will remain ($10^{-4} OH-).

Then use $K_w$ to determine the final concentration of H+.

  • $\begingroup$ Ok, I solved it and got for the H3O+ 10^10. And for pH =10.... But I have another question why is there H3O+ left if they all reacted with the OH- ? $\endgroup$
    – Mariam
    Feb 6, 2015 at 16:12
  • $\begingroup$ because aqueous solutions must obey the equillibrium Kw=[OH-][H+] = 10^-14. Even pure water dissociates to provide H+ and OH-. $\endgroup$
    – DavePhD
    Feb 6, 2015 at 16:16
  • $\begingroup$ Virtually all of the H3O+ ions reacted, but still very little amounts left because of water protolysis. $\endgroup$
    – Jeanno
    Feb 6, 2015 at 16:17

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