# Titration of barium(II) hydroxide by hydrochloric acid

$$\pu{25 ml}$$ of $$\pu{0.024 M}$$ $$\ce{HCl}$$ is titrated into a volumetric flask which initially contains $$\pu{45 ml}$$ of $$\pu{0.034 M}$$ $$\ce{Ba(OH)2}$$. Calculate the $$\mathrm{pH}$$.

This is how I calculated.

$$\ce{Ba(OH)2 + 2HCl -> BaCl2 + 2 H2O}$$

Moles leftover / excess moles will determine the $$\mathrm{pH}$$ because the others that have reacted will neutralise each other.

So calculating moles of $$\ce{Ba(OH)2}$$ and $$\ce{HCl}$$ in their respective volume $$C=\frac{n}{v}$$, hence $$n= c \times v$$.

Moles of $$\ce{Ba(OH)2} = \pu{0.034M} \times \frac{45}{1000} = 1.53 \times 10^{-3}$$

Moles of $$\ce{HCl} = 0.024 \times \frac{25}{1000} = 6 \times 10^{-4}$$ but $$2\times6\times10^{-4}$$ since it’s $$\ce{2HCl}$$ which is $$1.2 \times 10^{-3}$$.

Now to find leftover moles, you’ll have to subtract as such $${1.53} \times 10^{-3} - 1.2 \times 10^{-3} = 3.3 \times 10^{-4}$$, this is the excess moles of $$\ce{Ba(OH)2}$$.

Looking at solution of excess $$\ce{Ba(OH)2}$$

$$\ce{Ba(OH)2 -> Ba^2+ + 2 OH-}$$

Since $$\ce{Ba(OH)2}$$ : $$\ce{2OH-}$$, moles of $$\ce{OH-} = 3.3 \times 10^{-4} \times 2 = 6.6 \times 10^{-4}$$ moles of $$\ce{OH-}$$

So, concentration $$= \frac{6.6 \times 10^{-4}}{\frac{25+45}{1000}} = \frac{6.6 \times 10^{-4}}{0.07} = 0.009..$$

$$\mathrm{pOH} = -\log \ce{[OH]} = -\log[0.009] = 2.026$$

Since $$\mathrm{pH} + \mathrm{pOH} = 14$$, $$\mathrm{pH} = 14 - \mathrm{pOH}$$
So $$\mathrm{pH} = 14 - 2.026 = 11.97$$

HOWEVER, the book is saying $$\mathrm{pH} = 12.55$$. I have done this problem multiple times and cannot figure it out.

• >"Moles of $\ce{HCl} = 0.024 \times \frac{25}{1000} = 6 \times 10^{-4}$ but $2\times6\times10^{-4}$ since it’s $\ce{2HCl}$ which is $1.2 \times 10^{-3}$". The amount of reacted $\ce{Ba(OH)2} =\frac{\text{the amount of}~\ce{HCl}}{2}=\frac{6\times{10^{-4}}}{2}=\pu{3\times{10^{-4} mol}}$ – Adnan AL-Amleh Jun 27 at 5:38

Your solution is correct up to the point you assumed that you can double the concentration of hydrochloric acid. Unfortunately, this is wrong and not at all how stoichiometry works. Let's focus on what's important. In the nutshell, we are dealing with a typical neutralization reaction:

$$\ce{H3O+ + OH- <=> 2 H2O}$$

and note that $$\ce{BaCl2}$$ as a salt of a strong acid and a strong base won't undergo hydrolysis and, as a consequence, won't have a noticeable impact on $$\mathrm{pH}$$.

As you [semi-]correctly calculated, there is an excess of a base ($$n_0$$ values refer to the initial amounts):

\begin{align} n_0(\ce{H3O+}) &= n(\ce{HCl}) \\ &= c(\ce{HCl})\cdot V(\ce{HCl}) \\ &= \pu{2.4e-2 mol L-1}\cdot\pu{2.5e-2 L} \\ &= \pu{6e-4 mol} \end{align}

\begin{align} n_0(\ce{OH-}) &= 2\cdot n(\ce{Ba(OH)2}) \\ &= 2\cdot c(\ce{Ba(OH)2})\cdot V(\ce{Ba(OH)2}) \\ &= 2\cdot\pu{3.4e-2 mol L-1}\cdot\pu{4.5e-2 L} \\ &= \pu{3.06e-3 mol} \end{align}

One can see that there is a tremendous excess of hydroxide of the following amount:

\begin{align} n(\ce{OH-}) &= n_0(\ce{OH-}) - n_0(\ce{H3O+}) \\ &= \pu{3.06e-3 mol} - \pu{6e-4 mol} \\ &= \pu{2.46e-3 mol} \end{align}

and the leftover concentration is

\begin{align} c(\ce{OH-}) &= \frac{n(\ce{OH-})}{V(\ce{HCl}) + V(\ce{Ba(OH)2})} \\ &= \frac{\pu{2.46e-3 mol}}{\pu{2.5e-2 L} + \pu{4.5e-2 L}} \\ &= \pu{3.51e-2 mol L-1} \end{align}

Finally, as you rightfully suggested (use the correct value for $$c(\ce{OH-})$$, however):

\begin{align} \mathrm{pH} &= 14 - \mathrm{pOH} \\ &= 14 + \log(\pu{3.51e-2}) \\ &= 12.55 \end{align}

### Notes

• Using the term "moles of X" is incorrect. The proper terminology here is the amount of substance. Never try to substitute a physical quantity with its unit.

• Watch out for the proper capitalization, which is actually a big deal in natural sciences. $$\ce{HCl}$$ is a formula for hydrogen chloride, whereas $$\ce{HCL}$$ is not and may denote "hydrogen-carbon-ligand" or be an abbreviation. Same goes for math operators, including $$\mathrm{pH}$$ and $$\mathrm{pOH}$$: $$\ce{Ph}$$ is phenyl and $$\ce{PH}$$ can be read as "phosphor-hydrogen", or be yet another abbreviation.

• Thank you. I do have some questions. 1) why is it "H3O+ OH -> 2H2O" and not "2H + 2OH -> 2H2O" or are these equivalent? 2) Why did you multiply the moles of OH by 2? Is it because of Ba(OH)2 having 2*OH? 3) From where I went wrong and looking into your 1st bullet point in the notes, what topic should I look into? I don't exactly know what I don't know if that makes any sense. – soho Jun 25 at 21:52
• @soho 1. Yes, both are pretty much equal (check out Balancing redox equations in acidic media: to use H+ or H3O+?); 2. As it's mentioned, $\ce{Ba(OH)2}$ is a strong base: $$\ce{Ba(OH)2 <=>> Ba^2+ + 2 OH-}$$ 3. See the most recent Meta post Buzzwords Season 9 - Number of Moles (reboot) and have a look at the links inside. – andselisk Jun 26 at 10:05