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Question:

Calculate the pH produced from mixing $\pu{25.0 mL}$ of $\pu{0.420 M} $ $\ce{Ba(OH)_2}$ with $\pu{125 mL}$ of $\pu{0.120 M}$ $\ce{ HCl}$.

Attempt:
I'm learning about acids and bases right now (and not very used to it yet). First I wrote the chemical equation $$\ce{2HCl + Ba(OH)_2 \rightarrow 2H_2O + BaCl_2}$$

Then I calculated the number of moles of each $\ce{Ba(OH)_2}$ has $$ \pu{0.025 L} \times \pu{0.420 M} = 0.0105 $$ moles

and for $\ce{HCl}$ $$0.125 \pu{mL } \times \pu{0.120 M} =0.0150$$ So $\ce{Ba(OH)2}$ is the limiting reactant. However, I'm not sure how to proceed so any help would really be appreciated!

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You are on the right track just one thing. $\ce{Ba(OH)2}$ is in excess and $\ce{HCl}$ is the limiting reactant, which means all of the $\ce{HCl}$ is used up. That alone tells us the pH is above 7, as $\ce{Ba(OH)2}$ is alkaline and there is no acid left.

To find the exact pH of the resulting solution, first we need to find how much of $\ce{Ba(OH)2}$ is left. Since all of the $\ce{HCl}$ is used, we can find the number of moles of $\ce{Ba(OH)2}$ needed to react with it. As you've stated, the number of moles of $\ce{HCl}$ is 0.0150. From the equation we can see that 2 moles of $\ce{HCl}$ reacts with 1 mole of $\ce{Ba(OH)2}$. So 0.0150 moles of $\ce{HCl}$ will react with $\frac{0.0150}{2}$ = 0.0075 moles of $\ce{Ba(OH)2}$. But there are 0.0105 moles of $\ce{Ba(OH)2}$ present. So we find the difference to find how many moles there are in the solution when the reaction is finished.

This number is 0.0105 - 0.0075 = 0.003 moles. Now, one thing about acids and bases, is that their ions dissociate in solution. The more ions dissociate, the stronger the acid or base is. Here, there is no acid remaining, but only 0.003 moles of said base present. $\ce{Ba(OH)2}$ dissociates as follows: $$\ce{Ba(OH)2 <=> 2OH- + Ba^2+}$$ This means that one mole of $\ce{Ba(OH)2}$ dissociates to form 2 moles of $\ce{OH-}$ ions. So 0.003 moles of $\ce{Ba(OH)2}$ dissociates to form 0.003 * 2 = 0.006 moles of $\ce{OH-}$ ions. The resulting solution has a volume of 125 + 25 = 150 mL. The concentration of $\ce{OH-}$ ions is found by $\frac{0.006}{0.150}$ = 0.04 M.

To find the pH, it's easier to first find the pOH. That can be found using the formula: $$\ce{pOH = -log(OH^-)}$$ In your case, the value of $\ce{OH-}$ is 0.04 and the pOH is $$\ce{pOH = -log(0.04)}$$ or pOH = 1.397940. FINALLY, the pH can be found by subtracting the pOH from 14 because $$\ce{pOH + pH = 14}$$ In this case, pH = 14 - 1.397940 = 12.60206.

This is my first ever answer on this site, hope this helps. If I've made any mistakes, sorry and feel free to help out, ever-active community of Chemistry Stack Exchange

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  • $\begingroup$ In my opinion, it'd be easier to find $[H^{+}]$ by dividing $K_{w}$ by $[OH^{-}]$, and then finding the negative log of that. It'd skip a few steps of working with $pOH$ etc. $\endgroup$ – user60221 Apr 22 '18 at 9:09
  • $\begingroup$ It's all one and the same, really. That works too, but I personally like pOH. It' s like yin and yang, pH and pOH; two sides of a coin and whatnot. $\endgroup$ – iha99 Apr 22 '18 at 9:12
  • $\begingroup$ One thing though. If $Ba(OH)_{2}$ is in excess, then why does all of the $HCl$ react? Why doesn't all of the $Ba(OH)_{2}$ react to leave an excess of $HCl$? $\endgroup$ – user60221 Apr 22 '18 at 9:19
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    $\begingroup$ @iha99 Maybe this should be completely obvious, but where you say "You are on the right track. Ba(OH)2 is the limiting reactant, which means all of the HCl is used up. That alone tells us the pH is above 7, as Ba(OH)2 is alkaline and there is no acid left", I'm confused because if Ba(OH)2 is the limiting reactant, shouldn't it be used first and leave acid behind? $\endgroup$ – jjhh Apr 22 '18 at 9:22
  • $\begingroup$ Also, the total solution volume won't be 0.150 mol dm^-3. It will be 0.025 mol dm^-3 - you forgot to convert 0.125 mL to L. $\endgroup$ – user60221 Apr 22 '18 at 9:24

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