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I have the following question: Assume that you have two solutions. One of $50\ \mathrm{cm^3}$ $\ce{HCl}$ $\mathrm{pH} = 1.3$ and one of $150\ \mathrm{cm^3}$ $\ce{Ba(OH)2}$ $\mathrm{pH} = 12.3$. If we mix these two solutions what will the $\mathrm{pH}$ of the mixture be?

I have done the following:
$\mathrm{pH} = 1.3 \Rightarrow [\ce{H+}] = 10^{-1.3} = 0.05\ldots\ \mathrm M \Rightarrow n = 0.05\ldots \times 0.05 = 0.0025\ldots$
$\mathrm{pH} = 12.3 \Rightarrow \mathrm{pOH} = 1.7 \Rightarrow [\ce{OH-}] = 10^{-1.7} = 0.019\ldots \Rightarrow n = 0.019\ldots \times 0.15 = 0.0029\ldots$

My problem is I do not know how to proceed

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  • $\begingroup$ Please typeset your post. That will help others help you get the information you request. $\endgroup$ – Todd Minehardt Nov 6 '15 at 16:42
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The OH- and the H+ will react, neutralize each other to form H2O, H+ + OH- = H2O. Hence the ions remaining is: 0,0029... - 0,0025... = 0,00048... which will be OH- ions. I believe you can proceed on your own from here. Just calculate the concentration and then the pH.

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