Learning Goal: To calculate the pH at the equivalence point for various types of
ID: 877533 • Letter: L
Question
Learning Goal:
To calculate the pH at the equivalence point for various types of titrations.
The equivalence point in an acid-base titration is the point at which stoichiometrically equivalent quantities of acid and base have been mixed together. At this point the reaction is complete because all analyte has been consumed by titrant.
On a titration curve, the equivalence point is represented by the point of inflection (where the curve changes concavity).
The figure(Figure 1) shows the titration of 40.0 mL of 0.100 M HCl with 0.100 M NaOH. When 40.0 mL of the NaOH solution is added, the acid-base neutralization reaction is complete.
When analyzing titrations involving weak acids or bases, consider how the neutralization reaction will impact the pH of the system. For example, when titrating a weak acid with a strong base, at equilibrium all the bases has been used to neutralize the acid forming its conjugate weak base.
A. Match each type of titration to its pH at the equivalence point.
options: a. weak acid, strong base. b. strong base, strong base. c. weak base, strong acid.
go into boxes titled: PH less than 7. PH equal to 7. and PH greater than 7.
Part B
A 56.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 28.0 mL of KOH.
Part C
Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8×10?5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant.
Part D
A 30.0-mL volume of 0.50 M CH3COOH (Ka=1.8×10?5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH.
Explanation / Answer
Q. 1)
A 56.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 28.0 mL of KOH.
Here Both are strong (acid and base)
Lets write the reaction of acid and base
HBr + KOH --- > KBr + H2O
From the reaction we say that mol ratio of acid and base it 1:1
We find moles acid
Moles of acid (HBr) = Volume in L * Molarity = 0.056 L * 0.25 M = 0.014 mol
Moles of KOH = Volume in L* molarity = 0.028 L * 0.50 M = 0.014 mol
= 0.014 mol KOH
Now according to the balanced chemical reaction we say that moles of acid and base both reacted and salt is formed. So this solution is neutral and its pH is 7
Q. 2
Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8×105) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant.
We write the reaction:
NH3(aq) + HNO3 (aq) --- > NH4NO3 (aq)
We calculate moles each.
Moles of NH3 = Volume in L * molarity = 0.050 * 0.20 = 0.01mol NH3
Mol HNO3 = 0.050L * 0.20 M
= 0.01 mol
ICE chart
NH3(aq) + HNO3 (aq) --- > NH4NO3 (aq)
I 0.01 0.01 0
C -0.01 - 0.01 + 0.01
E 0 0 0.01
[NH4NO3]= 0.01 mol / total volume in L = 0.01 mol / (0.100 L )
= 0.1 M
NH4NO3 is the salt of the weak base and strong acid.
NH4NO3 dissociates completely
So [NH4NO3] = [ NH4+]
Lets show the reaction of NH4+ with water and ICE chart. Then we use kb expression to H3O+ concentration.
NH4+ (aq) + H2O (l) ------- > NH3 (aq) + H3O+ (aq)
I 0.1 M 0 0
C -x +x +x
E (0.1-x) x x
ka = [NH3] [H3O+] / [NH4+]
ka = 1.0 E-14 / kb = 1.0E-14 / 1.8E-5= 5.56 E-10
5.56 E -10 = x2 / ( 0.1-x)
Value of ka is very small so we use 5 % approximation.
5.56 E -10 = x2 / 0.1
5.56 E -11 = x2
x = 7.45 E-6
x = [H3O+] = 7.45 E-6 M
pH= -log [H3O+] = -log (7.45E-6) = 5.13
Q. 3
A 30.0-mL volume of 0.50 M CH3COOH (Ka=1.8×105) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH.
Mol CH3COOH = 0.030 L * 0.50 M =0.015 mol
Mol NaOH = 0.030 L *0.50 M = 0.015 mol
Reaction :
CH3COOH (aq) + NaOH (aq ) -------- > CH3COONa(aq) + H2O (l)
I 0.015 0.015 0
C -0.015 -0.015 +0.015
E 0 0 0.015
[CH3COONa]= [CH3COO-] = 0.015 mol / 0.060 L
=0.25 M
CH3COONa is the salt of strong base and weak acid. It will further dissociate and CH3COO- reacts with water.
CH3COO- (aq) + H2O (l) --- > CH3COOH (aq) + OH- (aq_
I 0.25 0 0
C -x +x +x
E ( 0.25-x) x x
Kb = 1.0E-14/ kb = [OH-] [CH3COOH] / [CH3COOH]
kb = 1.0E-14/ 1.8E-5 = x2/ (0.25-x)
kb = 5.56 E E-10 = x2/ (0.25-x)
since value of kb is very small we use 5 % percent approximation
5.56 E E-10 = x2/ 0.25
5.56 E E-10 * 0.25 = x2
x = 1.18 E -5
[OH]= 1.18 E-5 M
pOH = -log [OH-]
=-log (1.18E-5) = 4.93
pH = 14- pOH = 14- 4.93 = 9.07
pH = 9.07
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