Calculating the pH upon titrating barium hydroxide and hydrochloric acid

**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!




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*


 5