14.7 Acid Base Titrations

Acid – Base Titrations involve the calculations of the pH at different stages of the titration. The titration curves illustrate the plot of the pH against the titrant volume. The titration curves can illustrate the equivalence points of the titration as well as the behavior of the acid-base neutralization reaction during the continuous addition of the acid to the base or vice versa.

Reference: https://chem.libretexts.org/Courses/Brevard_College/CHE_104%3A_Principles_of_Chemistry_II/07%3A_Acid_and_Base_Equilibria/7.20%3A_Titration_Curves

There are several types of the titration curves:

1. Titration Curve: Strong Acid with Strong Base
2. Titration Curve: Strong Acid with Weak Base
3. Titration Curve Strong Base with Weak Acid
4. Titration Curve: Weak Acid with Weak Base

Typical titration curves for strong acid titrated against strong base or weak acid against strong base is given in the figure below:

Reference: https://www.khanacademy.org/test-prep/mcat/chemical-processes/titrations-and-solubility-equilibria/a/acid-base-titration-curves

Calculating pH for Titration Solutions: Strong Acid with Strong Base

10.0 mL of 0.0300 M HBr (aq) (the analyte) is titrated with 0.0100 M NaOH(aq)

solution (the titrant). Calculate the pH at each of the following points in the titration:

1. the initial point, before any titrant has been added;

2. after adding 15.0 mL of 0.0100MNaOH(aq)
3. after adding 30.0 mL of 0.0100MNaOH(aq)
4. after adding 45.0 mL of 0.0100MNaOH(aq)

;

Before carrying out any other calculations, determine:

1. the number of millimoles of analyte initially present, and
2. the volume of titrant that must be added to reach the equivalence point. It will be useful to keep these numbers in mind as we proceed to calculate the pH values throughout the course of the titration.

1. The number of millimoles of HBr initially present is

mmol HBr = (0.0300M) x (10.0mL) = 0.300 mmol

• The neutralization reaction for this titration is

HBr (aq)  +   NaOH (aq)    H₂O (l)  +  NaBr (aq)

At the equivalence point

mmol HBr initially present  =  mmol NaOH added

Therefore, the volume of NaOH (aq)

solution needed to reach the equivalence point is

V NaOH = M HBr x V HBr x M NaOH = 30.0 mL

Now, we are ready to carry out the calculations of the pH at the indicated points in the titration.

1.  The initial point, before any titrant has been added

Here we simply have a solution of 0.0300 M HBr. Being a strong acid, we can assume complete dissociation and

C HBr = 0.0300 M =  [H₃O ⁺]

pH= – log (0.0300) = 1.52

• After adding 15.0 mL of 0.0100 M NaOH (aq)

The number of millimoles of added NaOH is

mmol NaOH = (0.0100M) x (15.0mL) = 0.150 mmol

The total volume of the solution at this point is

Vt = V HBr + V NaOH = 10.0 mL+15.0 mL= 25.0 mL

We can summarize the numbers of millimoles of analyte and titrant before and after the addition by the reaction and modified ICE table

HBr (aq)   +   NaOH  (aq)      H₂O  (l)    +    NaBr  (aq)

Using ICE table:

HBr (aq)    +    NaOH  (aq)           H2O  (l)    +    NaBr  (aq)

Initial:                  0.300 mmol              0                                     0                            0

After RXN          – 0.150 mmol           0                                     0                     +0.150 mmol

Therefore,

[H₃O ⁺] =  C HBr = 0.150 mmol / 25.0mL= 6.00 × 10⁻³ M

the pH is:

pH =– log 6.00 × 10^–3 = 2.22

• After adding 30.0 mL of 0.0100 M NaOH (aq)

The number of millimoles of added NaOH is

mmol NaOH added = (0.0100M) x (30.0mL) = 0.300 mmol

This is the equivalence point, because we have added as many millimoles of NaOH (aq) titrant as there were millimoles of HBr analyte in the sample:

mmol NaOH added =  0.300 mmol  = mmol HBr initially present

Using ICE table:

HBr (aq)    +    NaOH  (aq)           H2O  (l)    +    NaBr  (aq)

Initial:                  0.300 mmol         0.300 mmol                     0                            0

After RXN           0                           0                                      0                     +0.300 mmol

The only source of H3O+ in the solution at this point is water’s autoprotolysis.  (Na+ and Br– from the dissociated NaBr

have no acid-base character.) Therefore,

pH =7.000

• After adding 45.0 mL of 0.0100MNaOH(aq)

This is 15.0 mL beyond the equivalence point. The number of millimoles of added NaOH is

mmol NaOH added = (0.0100 M) x (45.0mL) = 0.450 mmol

At this point our table of millimols of analyte and titrant before and after the addition now has the following values:

Using ICE table:

HBr (aq)    +    NaOH  (aq)            H2O  (l)    +    NaBr  (aq)

Initial:                  0.300 mmol         0.450 mmol                     0                            0

After RXN           0                           – 0.150 mmol                   0                     +0.450 mmol

The solution now contains 0.150 mmol

of NaOH beyond what was needed to neutralize the 0.300 mmol of HBr initially present. This excess OH−

ions are dissolved in the total volume of the solution at this point, which is the sum of the initial sample volume plus all the titrant volume we have added. The total volume of the solution at this point is thus

V total = V HBr + V NaOH = 10.0 mL + 4 5.0 mL  = 5 5.0 mL

Hence:

[OH^–] =  C NaOH = 0.150 mmol / 55.0 mL = 2.73 ×1 0^–3  M

pOH = – log 2.73 ×1 0^–3   = 2.56

pH = 14.0 – 2.56  = 11.44 = 11.4

Find the pH at each of the following points in the titration of 25 mL of 0.3 M HNO₂ with 0.3 M NaOH. The ka value is 4.50 × 10⁻⁴.

1. The initial pH
2. After adding 10 mL of 0.3 M NaOH
3. After adding 12.50 mL of 0.3 M NaOH
4. After adding 25 mL of 0.3 M NaOH
5. After adding 26 mL of 0.3 M NaOH

1. The initial pH

Since HNO₂ is a weak acid, the use of an  ICE table is required to find the pH. The question gives us the concentration of the HNO₂.

HNO₂ (aq)   +   H₂O  (l)    H₃O ⁺ (aq)   +    NO₂  ⁻ (aq)

Using the ICE Table

HNO2 (aq)   +   H2O  (l)    H3O + (aq)   +    NO2  − (aq)

Initial:                  0.3 M            –                                0                                0

Change:                – X                                                +X                           +X

Equilibrium:          0.3 M – X                                          +X                           +X

Ka =  [X] x {X]   /  [0.3 – X] = 4.50 × 10⁻⁴

Using the assumption method: 0.3 >>>> X

[X] ² =  [0.3 x 4.5×10⁻⁴] = 1.35 x 10⁻⁴

[X] =  √ 1.3.5 x 10⁻⁴ = 0.0116 = [H₃O ⁺ (aq)]

pH = 1.94

• After adding 10 mL of 0.3 M NaOH

The number of millimoles of HNO₂ to be neutralized is

(25mL) x  (0.3 mmol HNO₂ 1mL)  =   7.50   mmol  HNO₂

The number of millimoles of OH  ^– that will be added within 10 mL is:

(10 mL) x (0.3 mmol OH  ^– /1mL)  =  3  mmol  OH  ^–

To calculate the pH with this addition of base we must use an ICE Table:

Using the ICE Table

HNO2 (aq)   +   H2O  (l)     H3O + (aq)   +    NO2  − (aq)

Initial:                  7.5 mmol            –                         –                               0

Add:                        0                     3 mmol                –                               0

Change:                – 3 mmol          – 3 mmol              –                              + 3 mmol

Equilibrium:          4.5 mmol              0                           –                              + 3 mmol

However, this only gives us the millimoles. To get the concentration we must divide by the total volume. The total volume is the 25 mL original solution of HNO₂ plus the 10 mL of NaOH that was added. Therefore, the total volume is 25 mL + 10 mL = 35 mL

Concentration of HNO₂: 4.5 mmol  / 35 mL HNO₂ = 0.1287 M

Concentration of NO₂ ^–:  3 mmol  / 35 mL NO₂ ^–=  0.0857 M

The Henderson-Hasselbalch equation is used to obtain the pH because ratio of the conjugate base and k_a value and the ratio of the acid and k_a value must exceed 100 because the ratio of HNO₂ ^– to HNO₂ is 0.08570.1287= 0.666. This is between 0.10 and 10. The ratio of HNO₂ to k_a is 0.1287 M / 4.50 × 10⁻⁴ = 195 and the ratio of F^– to k_a is 0.0857 M / 4.5 × 10⁻⁴ = 286. This exceeds 100 and hence The Henderson-Hasselbalch equation is justified to be used.

pH =  pka  +  log  [A−] / [HA]

pH = −log (4.50 × 10⁻⁴)   +   log.  0.0857/ 0.1287

pH=3.17

• After adding 12.50 mL of 0.3 M NaOH

After adding 12.50 mL of 0.3 M NaOH

pH after adding 12.50 mL of 0.3 M NaOH is to be calculated.

The millimoles of OH^– added in 12.50 mL =  12.50 mL∗.3 mmol OH⁻ / mL = 3.75 mmol OH⁻

Using the ICE Table

To find the concentrations we must divide by the total volume. This is the initial volume of HNO₂, 25 mL, and the addition of NaOH, 12.50 mL. Therefore, the total volume is 25 mL + 12.50 mL = 37.50 mL

Concentration of HNO₂: 3.75 mmol HNO₂  / 37.50 mL = 0.100 M

Concentration of NO₂^–: 3.75 mmol NO₂  ^–  /  37.50 mL =  0.100M

This is half of the neutralization point and because the total amount of acid to be neutralized, 7.50 mmol, has been reduced to half of its value, 3.75 mmol. At the half-neutralization point we can simplify the Henderson-Hasselbalch equation and use it. Since the amount of conjugate base NO₂^–and acid HNO₂ are equal and each quals 0.1 M. Furthermore, we know that log 1 = 0

pH = pK + log [0.100 / 0.100] = pKa

pH= − log 4.50 x 10 ^-4 = -3.35

• After adding 25 mL of 0.3 M NaOH

pH is needed to be calculated after the addition of 25 mL of NaOH.

The millimoles of OH^– added in 25 mL= 25 mL∗.(0.3 mmol OH ⁻ / 1mL) =7.5 mmol OH⁻

Using the ICE Table

This is the equivalence point of the titration. We know this because the acid and base are both neutralized and neither is in excess. To find the concentrations we must divide by the total volume. This is the initial volume of HNO₂ of 25 mL, and the addition of NaOH, 25 mL. Therefore, the total volume is 25. mL + 25. mL = 50. mL

Concentration of NO₂ ^—: 7.5 mmol NO₂ ^—   /  50. mL  = 0.15 M

However, to get the pH at this point we must realize that NO₂ ^– will hydrolyze. An ICE table is needed:

Using the ICE Table

In this reaction the NO₂ ^– acts as a base and Kb value is used instead of the Ka value.

Kb =  Kw / Ka  =  [1.0 x 10 ^{– 14}]  / [4.5 x 10 ^– 4]

Kb =  1.0×10−146.6×10−4

Kb =  2.22 × 10  ^{− 11}

Now that we have the Kb value, we can calculated the pH using the ICE table information:

Kb =  2.22 × 10  ^{− 11} = 2.22  x 10  ^{– 11} = { [X] * [X] }  /  [0.15 – X]

Using the simplified assumption: 0.15 >>> X

2.22  x 10  ^{– 11} = { [X] * [X] }  /  [0.15]

[X] ²  =  [2.22  x 10  ^{– 11}] * [0.15] = 3.33 x 10 ^{– 12}

[X]  =  √ 3.33 x 10 ^{– 12}   =  1.83 x 10 ^{– 6}

pOH = – log [1.83 x 10 ^{– 6}  = 5.74

The pH = 14.0 – 5.74  =  8.26

• After adding 26 mL of 0.3 M NaOH

The pH is to be calculated after the addition of 26 mL of NaOH.

The millimoles of OH  ^— added in the 26 mL equal:

26 mL ∗.3 mmol OH  ^— / 1mL  = 7.8 mmol OH  ^—

This amount is greater than the moles of acid that is present. The 7.8 mmol OH  ^—  neutralizes the 7.50 mmol HNO₂. To find how much OH^– will be in excess we subtract the amount of acid and hydroxide.

mmoles of hydroxide in excess =  7.8 mmol  –  7.50  mmol  =   0.3 mmol OH  ^—

To find the concentration of the OH^– we must divide by the total volume. This is the initial volume of HNO₂, 25 mL, and the addition of NaOH, 26 mL. Therefore, the total volume is 25 mL + 26 mL = 51 mL

The concentration of OH  ^—  is 0.3  mmol OH  ^—  /  51 mL  =  0.00588 M

pOH  =  – log [0.00588]   =  2.23

pH =  14.0  –  2.23  =  11.77 = 11.8

 The videos below illustrate the concept of the titration and the titration curves:

Acid – Base Indicators

Indicators are organic dyes that are used to detect the pH of aqueous solutions. The general ionization equation of an indicator in aqueous solution is given below:

HIn  (aq)   +  H₂O  (l)        H₃O ⁺ (aq)       +       In ^– (aq)

Acid            Base                 Conjugate acid              Conjugate base

Ka  =  { [H₃O ⁺ (aq)]  *  [In ^– (aq)] }  /  [HIn]

HIn (aq) is the acidic form of the indicator.

In ^– (aq) is basic form of the indicator.

This is an equilibrium reaction, when acidic is added in a small amount, the equilibrium will shift to the left to the HIn acidic form of the indicator. On the other hand, when a small amount of a base is added, the equilibrium will shift to the right because the incoming base will scavenge the hydronium acid. The equilibrium will shift to the In ^– basic form of the indicator.

The Hassel Balch equation can be used to calculate the pH of the indicator and its composition:

Ka  =  { [H₃O ⁺ (aq)]  *  [In ^– (aq)] }  /  [HIn]

[H₃O ⁺ (aq)]  =  {Ka    *  { [HIn]  /  [In ^– (aq)] }

Taking the – log of both sides:

• log [H₃O ⁺ (aq)]  =  – log {Ka    *  { [HIn]  /  [In ^– (aq)] }

pKa  =  pKa   +  log  [[In ^– (aq)]   /  [HIn]

pKa  = pKa  log [Base] / [Acid]

The color change interval and sometimes is referred to as the pH interval is defined as an interval of the pH of indicator between pKa = -1 and pKa = +1 where the color change of the indicator is observed.

The figure below illustrates different indicators function at specific pH ranges:

Reference: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acid_and_Base_Indicators/Acid_and_Base_Indicators

The figure below illustrates the selection of an indicator for specific titrations. The selection considers that the color change can be observed easily. For example, in the titration of strong acid against strong base the three indicators (phenolphthalein, litmus paper and methyl orange) will yield sharper endpoints

 OpenStax eBook, 2^nd edition, chapter 14.

Figure 14 diagrams how several common pH indicators work along a titration curve.  For example, when titrating a strong acid against strong base the three indicators, (phenolphthalein, litmus paper and methyl orange, yield sharp endpoints and easily visible color changes