Finding pH - weak acid, strong base
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1. Find pH.
30cm^3 of 0.1mol/dm^3 CH3COOH mix with
10cm^3 of 0.1mol/dm^3 NaOH.
[ Ka of CH3COOH = 1.7 * 10^-5 ]
2. Calculate25cm^3 of 0.050mol/dm^3 ethanoic acid titrated with
0.100mol/dm^3 NaOH
(i) initial pH
(ii) final pH
(iii) pH at equivalence point
(iv) after 10.00cm^3 NaOH was added
(v) after 15cm^3 NaOH was added.
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Originally posted by secretliker:
1. Find pH.
30cm^3 of 0.1mol/dm^3 CH3COOH mix with
10cm^3 of 0.1mol/dm^3 NaOH.
[ Ka of CH3COOH = 1.7 * 10^-5 ]
2. Calculate25cm^3 of 0.050mol/dm^3 ethanoic acid titrated with
0.100mol/dm^3 NaOH
(i) initial pH
(ii) final pH
(iii) pH at equivalence point
(iv) after 10.00cm^3 NaOH was added
(v) after 15cm^3 NaOH was added.
This is a classic acid-base equilibria question, and (this very same question, though with varying molarities and volumes) is in fact one of my favourite classic questions that I frequently force... I mean kindly offer, my students to practice on.
Now, as Moderators of the Homework Forum, we're not here to do your homework, but are here to moderate, assist and support your understanding, learning and personal academic evolutionary process. As such, for the expressed purpose of helping you understand (as opposed to doing your homework for you and writing out a complete worked solution), I will be guiding you along with hints and tips on what you need to do to solve this (and all other) problems.
Q1.
Write out the ICF (Initial Change Final) table, in no. of moles. Note the final volume (ie. volume of acid + volume of alkali).
You will have a buffer system made of up the weak acid (ethanoic acid) and the conjugate base (ethanoate ions).
Using acid dissociation constant Ka = ([protons][conjugate base])/[acid], make molarity of protons the subject, and hence determine pH.
Q2.
(i)
Using acid dissociation constant, Ka, find molarity of protons, hence pH.
(ii)
Final pH would be when excess NaOH is added till there is no change in pH. This would eventually approximate (assuming large quantities of excess) the pH of the NaOH(aq) used. To determine the pH, given the molarity of OH- (which is the given molarity of NaOH), calculate pOH, hence obtain pH.
(iii)
Work out ICF table (in no. of moles for Initial Change Final) for equivalence point. Note the final volume (ie. volume of acid + volume of alkali). Use this final volume, as well as the no. of moles of the salt at Final, calculate molarity of salt. The salt is the product of a strong alkali (Na+ has no tendency to covalent bond OH- ions) with and a weak acid (CH3COO- has tendency to covalent bond with H+), hence sodium ethanoate is a basic salt (since OH- will exceed H+ in solution). Therefore, the next step is to calculate the pH of the solution resulting from the hydrolysis of the conjugate base, the ethanoate ion. Using (Ka)(Kb)=Kw, and assuming this is at room temperature (don't forget that Kw = 1x10^-14 *only* at room temperature! The dissociation of water molecules into protons and hydroxide ions is endothermic, so can you tell if the pH of neutral boiling water is greater or less than 7? How about pH of neutral melting ice?), calculate the value of Kb. Using this value of Kb, and the formula Kb = ([hydroxide ions][conjugate acid])/[base], work out the molarity of hydroxide ions, and hence pOH, and therefore pH.
(iv)
Write out the ICF (Initial Change Final) table, in no. of moles. Note the final volume (ie. volume of acid + volume of alkali).
You will have a buffer system made of up the weak acid (ethanoic acid) and the conjugate base (ethanoate ions).
Using acid dissociation constant Ka = ([protons][conjugate base])/[acid], make molarity of protons the subject, and hence determine pH.
(v)
Write out the ICF (Initial Change Final) table, in no. of moles. Note the final volume (ie. volume of acid + volume of alkali).You will have two species that are responsible for making the solution basic or alkaline : excess hydroxide ions and the basic salt sodium ethanoate. However, because OH- is a much stronger base than the CH3COO- ion (because the negative formal charge is delocalized by resonance over 2 electronegative oxygen atoms in ethanoate anion, compared to only 1 oxgyen atom in OH- ion; hence OH- ion is significantly more unstable and thus a significantly stronger base),we can safely neglect the hydrolysis of the ethanoate ions and calculate the pH of the solution using only the molarity of the OH- ions. From the molarity of hydroxide ions in the final solution (again, note the final volume correctly!), work out pOH, and hence pH.
Everyone Loves Chemistry! -
Originally posted by secretliker:
1. Find pH.
30cm^3 of 0.1mol/dm^3 CH3COOH mix with
10cm^3 of 0.1mol/dm^3 NaOH.
[ Ka of CH3COOH = 1.7 * 10^-5 ]
Hint for Q1 : 20cm3 of acid is in excess.
find the new concentration of acid after NaOH is added
use Ka equation solve for [H+]. U need to consider the concentration of [CH3COO-] formed also to use the Ka equation.
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Wow I like the long paragraphs though I don't get everything.
I'll type out my solution to Q1.
1. Find pH.
30cm^3 of 0.1mol/dm^3 CH3COOH mix with
10cm^3 of 0.1mol/dm^3 NaOH.
[ Ka of CH3COOH = 1.7 * 10^-5 ]
No. of moles of NaOH = 1*10^-3
No. of moles of CH3COOH = 3*10^-3
NaOH is the limiting reactant. 2*10^-3 moles of CH3COOH would be left.
1*10^-3 moles of CH3COONa would be formed.
A weak acid with it's sodium salt forms a buffer system?
CH3COOH <--> CH3COO- + H+
CH3COONa -> CH3COO- + Na+ (Ionic salt dissociates completely)
CH3COO- + H2O -> CH3COOH + OH-
Kb = [CH3COOH][OH-] / [CH3COO-]
[CH3COO-] = (2*10^-3) * (1000/40) = 0.05 (I think this is wrong)
(1*10^-14) / (1.7*10^-5) = [OH-]^2 / 0.05 (Assume at 298K)
[OH-] = 5.42326*10^-6
pH = 14 - (-log[OH-]) = 8.73 (But answer is 4.5)
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Originally posted by secretliker:
A weak acid with it's sodium salt forms a buffer system?
Yes that's right. The weak acid neutralizes any incoming hydroxide ions (by donating protons), while the congugate base (present in the salt) neutralizes any incoming protons (by accepting protons). Hence, pH is buffered. -
Originally posted by secretliker:
(But answer is 4.5)
The correct answer (based on the information you provided, including Ka value), to be slightly more accurate (at 'A' levels, you are required to give the final answer to 3 significant figures, unless otherwise specified by the question), is 4.47.
Solution :
From ICF table, Final (mol) is 2 x 10^-3 mol ethanoic acid, 0.0 mol sodium hydroxide, 1 x 10^-3 mol sodium ethanoate.
Volume is (30+10)/1000 = 0.04 dm3
Ka = ([H+][CH3COO-]) / [CH3COOH]
1.7 x 10^-5 = ([H+][(1x10^-3)/0.04] / [(2x10^-3)/0.04]
[H+] = 3.4 x 10^-5
pH = 4.47 (to 3 sig fig)
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Thanks UltimaOnline.
I need to be more careful on the concentrations of ions. =)