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Wize University Chemistry Textbook > Buffers and Titrations

Titration Problems

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Example: Solving for the pH at Equivalence Point

What is the pH at the equivalence point when 40.00 mL of 0.89 M Benzoic acid (Ka = 6.3 x 10-5) is titrated with 1.0 M NaOH?

Since we are told we are at equivalence point, we had to remember that means that the amount of titrant added is equal to the amount of analyte we initially had.
In our case, it means that the amount of NaOH added (titrant) is equal to the amount of benzoic acid (analyte) we initially had.

Using this information, we know @ equivalence point: moles of benzoic acid=moles of NaOH
Benzoic acid:
v=40mL
c=0.89M
Can solve for moles using n=cv
n=(0.89M)(0.04L)
n=0.0356 moles

NaOH:
c=1M
And now since we calculated moles of benzoic acid, we know that is equal to the moles of NaOH since we are at equivalence point.
n=0.0356 moles
Use n=cv to solve for volume of NaOH
n=cv
v=n/c
v=0.0356moles/1M
v=0.0356L

Now we can write out a chemical reaction @ equivalence point (equal moles of titrant and analyte react):


Benzoic acid + OH- --> Benzoate + H2O
0.0356 moles 0.0356 moles 0 0
-0.0356 -0.0356 +0.0356 +0.0356 moles
0 0 0.0356 moles 0.0356 moles

After this reaction, we only have benzoate (a weak base) and water leftover.
The pH will be determined by the weak base that is leftover.

Write a chemical reaction to show how the benzoate (weak base) that is leftover reacts:

Benzoate + H2O ⇌ Benzoic acid OH-

Instead of moles, since we now just have a weak base reacting with water (there is no strong substance), we have a reaction in equilibrium, so our ICE table needs to have concentrations.

We will then plug in our equilibrium concentrations into our Kb expression for benzoate to determine [OH-] and then pOH and then finally pH!

Solve for the concentration of benzoate:

c=?
n=0.0356 moles
v=must consider a total volume!
v=0.04L + 0.0356L
v=0.0756L
n=cv rearrange to solve for c:
c=n/v
c=0.0356moles/0.0756L
c=0.471M

Benzoate + H2O ⇌ Benzoic acid OH-
I 0.471M / 0 0
C -x / +x +x E 0.471 -x / x x

Now we can plug in concentration values from the "E" (equilibrium) row in the ICE table into the Kb expression for Benzoate. This will alow us to solve for x (x=[OH-])


Kb=[Benzoic Acid][OH−][Benzoate]Kb=\frac{\left[Benzoic\ Acid\right]\left[OH^-\right]}{\left[Benzoate\right]}

Kb=[x][x]0.471−xKb=\frac{\left[x\right]\left[x\right]}{0.471-x}

At this point we need to solve for Kb. The question gives us Ka=6.3x10-5

Kw=KaxKb
Kb=Kw/Ka
Kb=1x10-14/6.3x10-5
Kb=1.59x10-10

Plug Kb into the Kb expression:
1.59x10−10=[x2]0.471−x1.59x10^{-10}=\frac{\left[x^2\right]}{0.471-x}

Here we can check if y/K > 400 and we see that 0.471/1.59x10-10 > 400 so we can simplify:

1.59x10−10=x20.4711.59x10^{-10}=\frac{x^2}{0.471}

1.59x10-10(0.471)=x2
x=8.65x10-6 =[OH-] according to our ICE table!


Next we can solve for pOH:

pOH=-log[OH-]
pOH=-log[8.65x10-6]
pOH=5.06

Finally, we can solve for pH:

pH + pOH =14
pH=14 - pOH
pH=14-5.06
pH=8.94!!!!!!!

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Example: Everything You Could Be Asked About Titrations

A 200.0 mL sample of ammonia (Kb = 1.8 × 10-5) is titrated with 0.700 M hydrochloric acid. It requires 85.71 mL of hydrochloric acid to reach equivalence point.

a) What is the concentration of ammonia in the original sample?

The values given for H+ are values for reaching equivalence point and at equivalence point there are an equal number of moles of acid and base so we first have to calculate moles of H+, then we know that NH3 will have the same number of moles, and then using moles of NH3 we can calculate the concentration of it!

net reaction: NH3   +       H+       ⟶       NH4  +NH_3\ \ \ +\ \ \ \ \ \ \ H^+\ \ \ \ \ \ \ \longrightarrow\ \ \ \ \ \ \ NH_4^{\ \ +}
0.2000 L 0.08571 L
0.700 M
Use n=cv

n(H+)=(0.08571 L)(0.700 molL)(11)=0.0600 moln\left(H^+\right)=\left(0.08571\ L\right)\left(0.700\ \frac{mol}{L}\right)\left(\frac{1}{1}\right)=0.0600\ mol
This means n(NH3)=0.0600 since @ equivalence point the number of moles of acid and base are equivalent!

original sample has a volume of 200.0 mL of NH3. Use n=cv to solve for c.
c=n/v
[NH3]=0.0600 mol ÷0.2000 L  =0.300 molL\left[NH_3\right]=0.0600\ mol\ \div0.2000\ L\ \ =0.300\ \frac{mol}{L}


PAGE BREAK
A 200.0 mL sample of ammonia (Kb = 1.8 × 10-5) is titrated with 0.700 M hydrochloric acid. It requires 85.71 mL of hydrochloric acid to reach equivalence point.

b) What is the initial pH of the sample, before adding acid?

NH3 is a weak base in water – use an ICE table to determine the pH. (Note with just the NH3 in solution this is like a weak base problem, determine the pH of the buffer!!)

NH3    +     H2O   ⇋  NH4  +   +   OH−NH_3\ \ \ \ +\ \ \ \ \ H_2O\ \ \ \leftrightharpoons \ \ NH_4^{\ \ +}\ \ \ +\ \ \ OH^-

initial 0.300M __ 0 0
change -x __ +x +x
Eqm 0.300-x __ x x

Kb=[NH4  +][OH−][NH3]K_b=\frac{\left[NH_4^{\ \ +}\right]\left[OH^-\right]}{\left[NH_3\right]}

1.8×10−5=(x)(x)0.300−x    [NH3]>> KbNH3 so cross out ′′−x′′1.8\times10^{-5}=\frac{\left(x\right)\left(x\right)}{0.300-x}\ \ \ \ \left[NH_3\right]>>\ K_{bNH_3}\ so\ cross\ out\ ''-x''

x=[OH−]=2.3×10−3→ pOH=−log⁡(2.3×10−3)=2.63x=\left[OH^-\right]=2.3\times10^{-3}\rightarrow\ pOH=-\log\left(2.3\times10^{-3}\right)=2.63
pH=14−pOH=14−2.63=11.37pH=14-pOH=14-2.63=11.37


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A 200.0 mL sample of ammonia (Kb = 1.8 × 10-5) is titrated with 0.700 M hydrochloric acid. It requires 85.71 mL of hydrochloric acid to reach equivalence point.

c) What is the equivalence point pH?

At the equivalence point, all the NH3 has been protonated by HCl into NH4+. All the HCl added has been deprotonated by NH3 into Cl−.

NH3   +    H+    →   NH4  +NH_3\ \ \ +\ \ \ \ H^+\ \ \ \ \rightarrow\ \ \ NH_4^{\ \ +}
0.0600 mol 0.0600 mol 0
0 0 0.0600 mol
NH4 +   +H2O       ⇌      NH3          +H3O+NH_4^{\ +}\ \ \ +H_2O\ \ \ \ \ \ \ \rightleftharpoons \ \ \ \ \ \ NH_3\ \ \ \ \ \ \ \ \ \ +H_3O^+

Solve for [NH4]+ using n=cv
c=n/v
total volume = 200.0 mL + 85.71 mL = 285.7 mL

nNH4+=nNH3=0.0600 moln_{NH_4^+}=n_{NH_3}=0.0600\ mol

[NH4 +]=0.0600 mol  ÷ 0.2857 L=0.210 molL\left[NH_4^{\ +}\right]=0.0600\ mol\ \ \div\ 0.2857\ L=0.210\ \frac{mol}{L}

[NH4+]\left[NH_4^+\right] is a weak acid in water -use an ICE table to determine the pH.
 \
Ka NH4+=KwKb NH3=1.0×10−141.8×10−5=5.6×10−10K_{a\ NH4+}=\frac{K_w}{K_{b\ NH3}}=\frac{1.0\times10^{-14}}{1.8\times10^{-5}}=5.6\times10^{-10}

NH4 +   +H2O       ⇌      NH3          +H3O+NH_4^{\ +}\ \ \ +H_2O\ \ \ \ \ \ \ \rightleftharpoons \ \ \ \ \ \ NH_3\ \ \ \ \ \ \ \ \ \ +H_3O^+
Initial 0.210 ― 0 0

Change − x ― + x + x
Equilibrium 0.210 − x ― x x

Ka=[NH3][H3O+][NH4  +]K_a=\frac{\left[NH_3\right]\left[H_3O^+\right]}{\left[NH_4^{\ \ +}\right]}

5.6×10−10=(x)(x)0.210−x   [NH4  +]>> KaNH4+so cross oout ′′−x′′5.6\times10^{-10}=\frac{\left(x\right)\left(x\right)}{0.210-x}\ \ \ \left[NH_4^{\ \ +}\right]>>\ K_{aNH4+}so\ cross\ oout\ ''-x''

X=[H3O+]=1.1×10−5→  pH=−log⁡(1.1×10−5)=4.97X=\left[H_3O^+\right]=1.1\times10^{-5}\rightarrow\ \ pH=-\log\left(1.1\times10^{-5}\right)=4.97



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A 200.0 mL sample of ammonia (Kb = 1.8 × 10-5) is titrated with 0.700 M hydrochloric acid. It requires 85.71 mL of hydrochloric acid to reach equivalence point.

d) Solve for pH when 42.86mL of HCl has been added to the sample

hint: what point does this represent on a titration curve?
Notice that 42.86 mL is more or less half of 85.71 mL, i.e. 42.86 mL is the volume halfway to the equivalence point. The halfway to equivalence point is the pKa point, where pH = pKa. For bases, pOH = pKb.

pOH = pKb = − log Kb = − log (1.8 × 10-5) = 4.74
pH = 14 − pOH = 14 − 4.74 = 9.26

Alternate way to solve is using an ICE table:

NH3(aq)+ H3O(aq)+⇋  NH4(aq)++H2O(l)NH_{3\left(aq\right)}+\ H_3O_{\left(aq\right)}^+\leftrightharpoons\ \ NH_{4\left(aq\right)}^++H_2O_{\left(l\right)}
I 0.06 mol 0.03mol 0 /
C -0.03 -0.03 +0.03 /
E 0.03 0 0.03
We are left with a weak base and its conjugate acid (buffer system)
Can use the Henderson Hassalbalch equation to solve for pH;
HH: pH=pKa+ log[conj base]/conj acid]

We are given Kb -> pKb=-log(kb)=-log(1.8x10-5)=4.74
pKa+pKb=14
14-4.74=pKa
pKa=9.26

pH=9.26 +log[0.03 mol/0.03 mol)
pH=9.26

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A 200.0 mL sample of ammonia (Kb = 1.8 × 10-5) is titrated with 0.700 M hydrochloric acid. It requires 85.71 mL of hydrochloric acid to reach equivalence point.

e) What is the pH after 71.43 mL of hydrochloric acid has been added to the sample?



NH3      H+      Cl−      H2ONH_3\ \ \ \ \ \ H^+\ \ \ \ \ \ Cl^-\ \ \ \ \ \ H_2O
NH3   +    H+   ⟶    NH4  +NH_3\ \ \ +\ \ \ \ H^+\ \ \ \longrightarrow\ \ \ \ NH_4^{\ \ +}
0.0600 mol 0.07143 L 0
0.0700 M 0.0500mol
nH+n_{H+} = (0.700 mol/L) (0.07143 L) = 0.0500 mol

You have 0.0600 mol NH3NH_3 and 0.0500 mol H+, and NH3 and H+ react in a 1:1 mole ratio.
This is an excess/limiting question. The excess reagent is NH3 and the limiting is H+. The
0.0500 mol H+ reacts with 0.0500 mol NH3 to produce 0.0500 mol NH4+. There is an excess of 0.0100 mol NH3. In solution, you thus have 0.0100 mol NH3 and 0.0500 mol NH4 +−NH_4^{\ +}- a conjugate acid/base mixture.

pH=pKa+log⁡(NH3NH4)=−log⁡(5.6×10−10)+log⁡(0.0100mol0.0500mol)=8.56pH=pK_a+\log\left(\frac{NH_3}{NH_4}\right)=-\log\left(5.6\times10^{-10}\right)+\log\left(\frac{0.0100mol}{0.0500mol}\right)=8.56


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A 200.0 mL sample of ammonia (Kb = 1.8 × 10-5) is titrated with 0.700 M hydrochloric acid. It requires 85.71 mL of hydrochloric acid to reach equivalence point.


f) What is the pH after 90mL of HCl has been added?



NH3       +H+    ⟶    NH4 +NH_3\ \ \ \ \ \ \ +H^+\ \ \ \ \longrightarrow\ \ \ \ NH_4^{\ +}
0.0600 mol 0.09000 L
0.700 M

nH+n_{H+} = (0.700 mol/L) (0.09000 L) = 0.0630 mol


total volume = 200.0 mL + 90.00 mL = 290.0 mL

You have 0.0600 mol NH3 and 0.0630 mol H+, and NH3 and H+ react in a 1:1 mole ratio.
This is an excess/limiting question. The limiting reagent is NH3 and the excess is H+. The 63.0 mmol H+ reacts with 60.0 mmol NH3 to produce 60 mmol NH4 + .There is an excess of 3.0 mmol H+.

Now calculate the concentration of the H3O+(n=cv, c=n/v)
[H3O+]=0.0030 mol ÷0.2900 L =0.010 molL\left[H_3O^+\right]=0.0030\ mol\ \div0.2900\ L\ =0.010\ \frac{mol}{L}
pH = − log [H3O+] = − log (0.010) = 1.99



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A 200.0 mL sample of ammonia (Kb = 1.8 × 10-5) is titrated with 0.700 M hydrochloric acid. It requires 85.71 mL of hydrochloric acid to reach equivalence point.

g) Draw the titration curve



Note that the pH of the equivalence point must be <7 because a strong acid reacted with a weak base (strong acid is stronger so pH is lower @ equivalence point)
The weak base is titrated with the acid so it means we start with weak base (hence the pH ~11 to start with, and slowly start adding strong acid (strong acid added along the x axis)
Finally, at the bottom right of the curve, we have added so much acid that the pH drops to very low ~1.
Extra Practice
Calculate the pH of the solution resulting from the addition of 10.0 mL of a 0.10 M NaOH to 50.0 mL of 0.10 M HCN (Ka = 4.9 x 10-10) solution
Acids and Bases: Titrations
25mL of a solution of 0.5mol/L KOH is required to neutralize 15mL of sulphuric acid. What is the concentration of the acid?
2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(ℓ)2KOH\left(aq\right)+H_2SO_4(aq)\to K_2SO_4(aq)+2H_2O(\ell)
Henderson Hasselbalch: Titrations and pH
For the titration of 50.00 mL of 0.1000 M ammonia (NH3) with 0.1000 M HCl calculate the pH (For ammonia Kb​=1.8×10−5)(For\ ammonia\ Kb​=1.8\times10^{-5})
Before the addition of any HCl
After 20.00 mL of the acid has been added
Redox
A 5.000 mL sample of a solution of hydrogen peroxide, H2O2, was analyzed by titration with potassium permanganate, KMnO4, according to the reaction shown below. The H2O2 sample required 42.8 mL of 0.0175 mol L−1 KMnO4 (aq) to reach the end point.
2 KMnO4 + 5 H2O2 + 6 H+ → 5 O2 + 2 Mn2+ + 2 K+ + 8 H2O
What was the concentration of the H2O2 solution? Choose the closest value.
To a 100 mL of 0.1M of CH3COOH\rm CH_3COOH (acetic acid) is added 25mL of 0.17M NaOH\rm NaOH. The Ka\rm K_a of acetic acid is 1.8×10−51.8 \times 10^{-5}. Calculate the pH\rm pH of the solution. Please report your answer to two decimal places.
25L of a solution of 0.5M KOH is required to neutralize 15L of sulphuric acid. What is the concentration of the acid? Balance the equation if it is unbalanced.
KOH + H2SO4 --> K2SO4 + 2H2O
What volume of hydrochloric acid of 0.1M is required to neutralize 500mL at 0.5M sodium hydroxide?
HCl + NaOH --> NaCl + H2O
Titration Problem
We titrated 40mL of 0.2M HBr with 0.4M NaOH...
The pH meter indicated that the initial pH was 2.9, pH at the half equivalence point was 3.5, equivalence pt was at pH 8.9.
a) What is Ka for HBr?
What volume of 3.16 M H3PO4 solution is necessary to react with 27 mL of 0.18 M Mg(OH)2?
Acid Base Calculation
An impure sample of solid Na2CO3 was reacted with 0.1755 M HCl. A 0.2337 g solid sample required 15.55 mL of HCl solution.
Practice: Long Titration Problem
For the titration of 50.00 mL of 0.1000 M ammonia (NH3) with 0.1000 M HCl calculate the pH
Before the addition of any HCl
After 20.00 mL of the acid has been added
Acids and Bases: Strong Acid Mixture
Practice Question: Strong Base Mixture
Calculate the pH and pOH of a solution containing a mixture of 75.00 mL of 0.100 M HI and 25.00 mL of 0.250 M HClO4.
Buffers: Titration of Sodium Citrate
A student dissolves a sodium citrate (Na3C6H5O7; MW = 258.069 g/mol) tablet in 350.00 mL of water. The student then pipetted 20.00 mL of the solution and performed a titration using a stock solution of HCl previously standardized to a concentration of 0.9899 M. The experiment required 33.55 mL of the HCl solution to reach the endpoint and fully neutralize the sodium citrate solution. Calculate the concentration of the sodium citrate solution.
Titrations
A student prepares a potassium hydrogen phosphate (K2HPO4, MW = 156.1739 g/mol) solution and pipettes 20.00 mL of this solution for a titration. In the titration, it required 23.43 mL of 0.100 M HCl to reach the endpoint. What is the concentration of the potassium hydrogen phosphate solution?
Strong Acids and Bases
Calculate the pH of a basic solution prepared by taking 10.00 mL of 0.700 M NaOH solution and diluting it to 100.00 mL with water.
Titrations
A student prepares a potassium hydrogen phosphate (K2HPO4, MW = 156.1739 g/mol) solution and pipettes 20.00 mL of this solution for a titration. In the titration, it required 17.51 mL of 0.09977 M HCl to reach the endpoint. What is the concentration of the potassium hydrogen phosphate solution?
pH: Neutralization
A student adds 11.6 mL of a 3.7M KOH solution, to 29 mL of a 1.4 M HCl. What is the final pH after the addition?
Calculate the pH of the solution resulting from the addition of 10.0 mL of a 0.10 M NaOH to 50.0 mL of 0.10 M HCN (Ka = 4.9 x 10-10) solution
Chemical Reactions: Limiting Reagent
15mL of a 1.2M solution of HCl reacts with 22mL of a 1.4M solution of NaOH. How many moles of NaOH are left after the reaction is complete?
HCl(aq)+NaOH(aq)→H2O(l)+NaCl(aq)HCl_{(aq)}+NaOH_{(aq)}\rightarrow H_2O_{(l)}+ NaCl_{(aq)}
For the titration of 50.00 mL of 0.1000 M ammonia (NH3) with 0.1000 M HCl calculate the pH. Enter 2 decimal places with all answers. Kb=1.8x10-5
Before the addition of any HCl
After 20.00 mL of the acid has been added
Acids and Bases: Titrations
25L of a solution of 0.5M KOH is required to neutralize 15L of sulphuric acid. What is the concentration of the acid? Balance the equation if it is unbalanced.
KOH + H2SO4 --> K2SO4 + 2H2O
Acids and Bases: Titrations
What volume of hydrochloric acid of 0.1M is required to neutralize 500mL at 0.5M sodium hydroxide?
HCl + NaOH --> NaCl + H2O
For which type of titration will the pH be basic at the equivalence point?
A student titrated 10.00 mL of 0.10 M sodium acetate (CH3COONa) with HCℓ and determined that 10.00 mL of HCℓ was needed to reach the equivalence point. At equivalence, what is the expected pH?
Ka of CH3COOH = 1.8×10-5
A student titrated 10.00 mL of 0.10 M sodium acetate (CH3COONa) with HCℓ and determined that 10.00 mL of HCℓ was needed to reach the equivalence point. At equivalence, what is/are the major species in solution (besides water and any pH-neutral ions)?
Calculate the pH of the solution resulting from the addition of 10.0 mL of a 0.10 M NaOH to 50.0 mL of 0.10 M HCN (Ka = 4.9 x 10-10) solution
Henderson Hasselbalch: Titrations
Suppose you performed a potentiometric titration on the solutions in flasks A, B, and C by a gradual addition of HCℓ (aq) to each flask shown.