pH (1 Viewer)

Aerath · Aug 22, 2009

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10mL of 0.5M HCl acid is mixed with 25mL of 0.35M sodium hydroxide. Find the pH of the final product, assuming salt is neutral.
Thanks.

thisnameistaken · Aug 22, 2009

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2.[FONT=&quot] [/FONT]HCl<sub> (aq)</sub> + NaOH<sub> (aq)</sub> <sub><!--[if gte vml 1]><v:shapetype id="_x0000_t75" coordsize="21600,21600" o:spt="75" o

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</sub> NaCl<sub> (aq)</sub> + H<sub>2</sub>O<sub> (l)</sub>
3.[FONT=&quot] [/FONT]As can be seen, the molar ratio of reaction for the acid and base is 1:1.
4.[FONT=&quot] [/FONT]Calculating the number of moles of acid and base, using n = c × v:
5.[FONT=&quot] [/FONT]Acid: n = 0.5 × 0.01 = 0.005
6.[FONT=&quot] [/FONT]Base: n = 0.35 × 0.025 = 0.00875
7.[FONT=&quot] [/FONT]There are more moles of base than there are acid; hence there will be an excess of base after neutralisation completes.
8.[FONT=&quot] [/FONT]Moles of base remaining = 0.00875 – 0.005 = 0.00375
9.[FONT=&quot] [/FONT]As NaOH only releases one OH<sup>-</sup> ion, [OH<sup>-</sup>] = 0.00375
10.[FONT=&quot] [/FONT] pOH = -log<sub>10</sub>(0.00375) = 2.426
11.[FONT=&quot] [/FONT]Hence pH = 14 – pOH = 11.6
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annabackwards · Aug 22, 2009

HCl + NaOH --> NaCl + H2O

n(NaOH) = cV = 0.35 x (25/1000) = 8.75 x 10^-3 = 0.00875 moles of the acid is needed to neutralise the base

n(HCl) = cV = 0.5 x (10/1000) = 0.005 moles

Therefore there's 0.00875 - 0.005 = 0.00375 moles of extra NaOH

So therefore there's 0.00375 moles of OH ions floating around in the final solution

EDIT: Therefore [OH-] = n/V = 0.00375/0.035 = 3/28

Now [H+] [OH-] = 10^-14 so [H+] = 10^-14 / (3/28) = 9.3333 x 10^-4

Therefore pH = -log [H+] = -log (9.3333 x 10^-4)
= 13.03

Aerath · Aug 22, 2009

What about this one?
20mL of 0.08M HCl is mixed with 30mL of 0.05M NaOH. What is the pH of the resultant solution?

Sorry, there's just this one bit I don't get.

kaz1 · Aug 22, 2009

HCl + NaOH --> NaCl + H₂O

n[HCl]= 0.08x0.02=1.6x10^-3
n[NaOH]= 0.03x0.05=1.5x10^-3
From this we can infer that there is 0.1x10^-3 mol of hydrogen ions since the other H⁺ ions reacted with the OH^-.

Total volume= 0.05L
C[H⁺]=0.1x10^-3/0.05=2x10^-3 mol/L
-log(2x10^-3)=2.7

astroe · Aug 22, 2009

Loooll.
I don't get any of this either.

/ fail chemistry.

Aerath · Aug 23, 2009

kaz1 said:
HCl + NaOH --> NaCl + H₂O

n[HCl]= 0.08x0.02=1.6x10^-3
n[NaOH]= 0.03x0.05=1.5x10^-3
From this we can infer that there is 0.1x10^-3 mol of hydrogen ions since the other H⁺ ions reacted with the OH^-.

Total volume= 0.05L
C[H⁺]=0.1x10^-3/0.05=2x10^-3 mol/L
-log(2x10^-3)=2.7

OK, so how come in this question you had to divide by the volume, whereas in the question above, you didn't, to get the right answer?

annabackwards · Aug 23, 2009

Aerath said:
OK, so how come in this question you had to divide by the volume, whereas in the question above, you didn't, to get the right answer?

Oh damn, i forgot about that. You're supposed to do it in the first example too, because pH is always based on the concentration of H+ ions in solution. The answer comes out to be a bit different though, is the answer right?

So i've edited my first post to show you what the difference should be

Aquawhite · Aug 23, 2009

Hehe, and concentration was learnt all the way back in preliminary... Anna >.> lol. You still would have went very well in your exams!

Aerath · Aug 23, 2009

annabackwards said:
Oh damn, i forgot about that. You're supposed to do it in the first example too, because pH is always based on the concentration of H+ ions in solution. The answer comes out to be a bit different though, is the answer right?

So i've edited my first post to show you what the difference should be

Funnily enough, the answer to both questions was correct [according to the solutions lol]. First answer was 11.6 (which I got wrong, I got 13.0), and the second was 2.7. But both used different 'methodologies' to get to the correct answer.

annabackwards · Aug 23, 2009

Aquawhite said:
Hehe, and concentration was learnt all the way back in preliminary... Anna >.> lol. You still would have went very well in your exams!

Aha, i can be forgiven because my trials are over for chem ^^

Aerath said:
Funnily enough, the answer to both questions was correct [according to the solutions lol]. First answer was 11.6 (which I got wrong, I got 13.0), and the second was 2.7. But both used different 'methodologies' to get to the correct answer.

Oh i see. Well the 1st answer is dodgy then

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