EXPERIMENT ONE pH TITRATIONS prelab work
1.
In each each of the the foll follow owin ing g sol solut utio ions ns,, one one of of the the ions ions is a wea weak k aci acid d or or bas base. e. Write the balanced net ionic equation for the acid/base equilibrium when this ion reacts with water (‘hydrolysis’ of the ion). ammonium chloride dissolved in water
NH4+ + H2O
→
NH3 + H3O+
sodium acetate dissolved in water
CH3CO2− + H2O
2.
→
CH3COOH + OH−
The pH of 0.100M NaCN is 11.2. a)
Writ Wr itee a net ioni ionicc equ equat atio ion n for for the the aci acid/ d/ba base se equi equili libr briu ium m
b)
Calculate [OH −] in the solution.
c)
Calculate the % hydrolysis of the CN – ion in this solution.
a)
In a solution of NaCN, the CN− ion hydrolyses as follows:
CN− + H2O b)
→
HCN + OH−
sinc ince the the pH is 11.2, 1.2, pOH pOH must ust be be 14 - 11. 11.2 2 = 2.8 2.8 hence [OH−] = 10−2.8 = 1.6× 10−3M
c)
3.
% hydrolysis = 100× [OH−]eqlm / [CN−]initial = 100 × 1.6 × 10−3 / 0.100 = 1.6%
A kno known wn vol volum umee (1.0 (1.00 0 mL) mL) of a 0.1 0.100 00M M NaOH NaOH sol solut utio ion n was was dil dilut uted ed to to a tot total al vol volum umee of 10. 10.0 0 mL (Solution A). Next 1.00 mL of this solution solution (Solution A) was diluted to a total volume volume of 50.0 mL (Solution B). B). What is the pH of solution solution B?
[OH−] in solution A is 1/10 × 0.100M = 0.0100M This is further diluted by a factor of 50 to obtain solution B, giving [OH−] = 1/50 × 0.0100M = 0.000200M = 2.00× 10−4 M pOH = log10[OH−] = 3.7;
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so pH = 14 − pOH = 10.3
chemistry2
physical experiment 3. page 1
14
4.
14
What is is the di difference be between the two pH titration experiments A and B?
H p
In experiment A, (strong) acid from a burette is added to (strong) base in a flask. In experiment B, (strong) base from a burette is added to (strong) acid in a flask.
A
7
H p
B 7
0
0 0
5
0
10 1 5 20 20 2 5 30 30 3 5 40 40 4 5 5 0
5
10 1 5 20 20
25 30 30 3 5 40 40 4 5 50 50
volume added
volume added
14
5.
13
The The dia diagr gram am sho shows the the pH pH tit titra rati tion on of a sol solut utio ion n of of aci acid d
12
HA (25.00 mL) with 0.120 M NaOH.
11
a)
Mark ark th the eq equiv uivalen alence ce poin oint on on th the tit titra rati tio on cur curv ve.
b)
What is the NaOH volume
10 9 8
21 mL
H 7 p
at the equivalence point?
6 5
Calculate the concentration of HA from
4 3
this volume and the other data supplied.
2 1
number of moles NaOH in a titration = 0.120 ∴
×
21/1000 = 02.52
moles HA in a titration (ie in 25mL of solution) = 2.52
×
10-2 moles
10-2 moles
×
vol NaOH added (mL)
-2
∴[HA]
= 2.52 × 10 /0.025L = 0.101 M
c)
What What is the the val value ue of the the pH pH at at the the equ equiv ival alen ence ce poin point? t?
d)
What can you dedu educe abo about the aci acid?
8.6
The acid is weak, since pH at the equivalence point is basic. (more insightfully, the pKa is 5.0, which is the pH at the half equivalence volume)
6.
List Listed ed are are the the colo colour urss of the the ind indic icat ator or brom bromoc ocre reso soll green in aqueous solutions of various pH values.
pH
Colour
2.5
yellow
3.5
yellow
5.5
blue
mostly HInd
All indicators are weak acids, where the acid (HInd) and its conjugate base (Ind −) have different colours. Complete the following table for bromocresol green.
7.0
blue
8.5
blue
HInd Colour
Ind – colour
Approximate pKa
yellow
blue
4.5
When the pH = pK a, equal amounts of HInd and Ind − are present, and the xp3 mostly mostly HInd colour is 2c-05a halfway between yellow and Ind blue. −
chemistry2
mostly Ind−
When the pH = pK a, equal amounts of HInd and Ind− are present, and the colour is halfway between yellow and blue.
physical experiment 3. page 2
Calculations:
The pH at the equivalence point for each titration is determined from the titration curve, not the indicator colour. The volume at the equivalence point can can then be estimated estimated from the curve, or from the raw data. POSTLAB WORK Incorporate your answers to these questions into the Discussion section of your report.
1.
With With the the aid aid of of an equ equat atio ion, n, acc accou ount nt for for the the fact fact tha thatt the the equi equiva vale lenc ncee poin pointt for the the acet acetic ic aci acid dsodium hydroxide titration does not occur at a neutral pH.
At the equivalence point, species present are Na+ and CH3CO2−. Na+ is neutral, not being a conjugate of a weak acid, so do not react with water. However, CH3CO2 is a weak base, so sodium acetate solution is basic. CH3CO2− + H2O
2.
CH3COOH + OH−
With With the the aid aid of of an equ equat atio ion, n, acc accou ount nt for for the the fact fact tha thatt the the equi equiva vale lenc ncee poin pointt for the the hyd hydro roch chlo loric ric acid - ammonia titration does not occur at a neutral pH.
At the equivalence point, species present are NH4+ and Cl−. Cl- is not basic, but NH4+ is a weak acid, so ammonium chloride solution is acidic. NH4+ + H2O
3.
NH3 + H3O+
From From the the pH pH rang ranges es ove overr which which the the colo colour urss of the the thr three ee indi indica cato tors rs cha chang nged ed,, esti estima mate te the the pK a of each indicator. Explain your answers answers in terms of of the acid/base equilibrium of each indicator.
See prelab question 7. pH = pKa of the indicator when [HInd] = [Ind−]. The colour change normally occurs in the pH range pKa ± 1. When pH = pKa −1 , the ratio is 90% HInd :10% Ind−, and the HInd colour should dominate. When pH = pKa +1 , the ratio is 10% HInd :90% Ind−, and the Ind− colour should dominate.
2c-05a xp3
chemistry2
physical experiment 3. page 3
4.
Using Using as a guid guidee the the titra titratio tion n curve curvess you you obtain obtained ed in this this expe experim riment ent,, expla explain in wheth whether er or or not not methyl red (pK a = 5.1) would be a suitable indicator for each of the following titrations: a)
0.10M HCl versus 0.10M KOH
14
Yes, the vertical part of the titration curve extends well beyond 5.1. Compare with the student titration curve for HCl / NaOH
H p
7
methyl red gives very small error in titration volume
5.1
b)
0
0.10M CH 3COOH versus 0.10M KOH
0
5
1 0 1 5 20 20 2 5 30 30 3 5 4 0 45 45 5 0
volume adde
No, the equivalence point for a typical weak acid / strong base is on the basic side of neutral, and far from the pKa of methyl red. Compare with student titration curve for acetic acid / NaOH
c)
0.10M HNO3 versus 0.10M NH 3.
Yes, the equivalence point for a typical strong acid / weak base is on the acidic side of neutral. Compare with the student titration curve curve for HCl / NH3
pH = − log [H+].
5.
Could solutions of pH > 7 be prepared by continued dilution of a 0.10M 0.10M
HCl solution at 25 °C? Explain your answer.
No, there will always be more H+ than OH−. In pure water, [H+] = [OH−], so any added acid will make [H+] > [OH−]. To calculate pH for (say) 1× 10−8M HCl, the effect of added H+ on the dissociation of water must be considered. 6.
This This que quest stio ion n rela relate tess to expe experim riment ent 2, 2, the the titr titrati ation on of of acet acetic ic acid acid solu soluti tion on wit with h NaOH NaOH a)
K a for acetic acid is 1.74
×
10−5. Calculate its pK a.
pH = −log10Ka = 4.76 b) b)
Expl Explai ain n the the rela relati tio onshi nship p betwe etween en [CH [CH3COOH] and [CH3CO2−] when, in experiment 2, exactly HALF the equivalence volume of NaOH solution has been added. Read from your titration curve, the pH at this stage of the titration.
equal,, since exactly half of the acetic acid has reacted with OH− equal c)
Briefly explain why the pK a in (a) and the pH in (b) should be similar.
(hint: substitute into the K a or Hendersen-Hasselbalch expression)
pH = pKa when [HA] = [A−], from the Henderson-Hasselbach equation equation pH = pKa - log([HA]/[A−])
2c-05a xp3
chemistry2
physical experiment 3. page 4
experiment experiment 1
pH TITRATIONS mark /20 then divide by 2 for mark /10 (minimum mark =3) =3) PREWORK Subtract 2 if not done, 1 or 2 if poor effort (??)
EXPERIMENTAL ACCURACY (10) Mark the precision of the correctly calculated results according to the tables below. The spreadsheet chem2-08 x1 spreadsheet.xls may be useful here.
[HCl]
M
=
M
M
NaOH % error mark 0–½ 4 ½–1 3 1–2 2 2–5 1 >5 0
M
CH3COOH % error mark 0–1 4 1–1½ 3 1½–2 2 2–5 1 >5 0
NH3 % error mark 0–3 2 3–5 1½ 5–10 1 10–15 ½ >15 0
pH TITRATION CURVES (penalties) -1 for wrong axes -1 for no data points, or points badly plotted -1 for each poor choice of equivalence volume (max -2) CALCULATIONS (penalties) –3 for each type of conceptual error –2 for each type of arithmetic error (check calculations*) –1 for sig figure error PRESENTATION (1) 1 mark for neat presentation of pH curves, data, and calculations DISCUSSION / QUESTIONS (9 marks as follows) question 1 2 3 4 5 6
marks 1 1 2 2 1 2
(9)
1 for sensible pK a’s, 1 for explanation explanation 1 for sensible answers, 1 for explanation based on experimental curves dilution of HCl 1/2 mark for each of (a) and (b); (b); 1 mark for (c)
useful pKa’s: methyl red 5.1 2c-05a xp3
phenolphthalein 9.6
bromophenol blue 4.2 chemistry2
physical experiment 3. page 5
