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Will someone help me figure out how can I solve the following question?

We are given 110 ml of acetic acid (pKa=4.76), with concentration of 1M, to which we add 10 ml of NaOH with concentration of 1M.

Afterwards, we add the solution to 880 ml of distilled water.

What would be the pH of the solution, after the addition of water?

Will you please help me figure this out?

I obviously need to use Henderson-Hasselbalch here, but how does the addition of water change things?

Thanks a lot in advance

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closed as off-topic by Wildcat, bon, Jan, Todd Minehardt, Klaus-Dieter Warzecha Oct 1 '15 at 14:23

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Perhaps I can help guide you through this problem.

First, what is the reaction that occurs between acetic acid and NaOH? Write it down and check that it is balanced. What are the reactants and what are the products.

Next, using the stoichiometry of the reaction figure out how much reactants and products you have at the end of the reaction (is the limiting reactant used to completion?)

Then you can figure out the molarity of your reactants and products (keep in mind the addition of water will change your final volume)

Finally, enter the molarity of your acid and conjugate base into the H-H equation and solve for pH.

Happy to help you work through it if you get stuck.

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So I presume you can work out the initial H-H equation. I got pH 5.8? might be worth checking. Then use the equation $$c_1 \cdot V_1 = c_2 \cdot V_2$$ Where $c_1 = [\ce{H+}]$

$[\ce{H+}] = 10^{-\mathrm{pH}}$

so I got $[\ce{H+}] =10^{-5.8} = 1.58489 \times 10^{-6}$

Now we can rearrange our original equation: $(c_1 \cdot V_1)/V_2 = c_2$

$c_1=1.585\times10^{-6}$

$V_1=120\ \mathrm{ml}=0.12\ \mathrm L$

$V_2=(120\ \mathrm{ml}+880\ \mathrm{ml})= 1\ \mathrm L$

$c_2=1.902\times10^{-7}$

$\mathrm{pH}=-\log[\ce{H+}]$

$\mathrm{pH}= -\log(1.902\times10^{-7})$

$\mathrm{pH}= 6.72$

Hope this helps, Its been a long time since I did anything like this so let me know how it goes and if this is correct. cheers.

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