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Though the pH at various points during the titration of a strong acid with a strong base (or vice versa) is relatively easy to determine, pre-AP and AP Chemistry

teachers alike would be wise to apply the ICF technique early as it will build student confidence for its application to more difficult problems later. Consider the curve that describes the titration of 25.0 mL of 0.100 mol/L HCl with NaOH.

FIGuRE 1

pH

Providing such a curve for the titration of 25 mL of 0.100 mol/L HCl, with no further information, the AP Chemistry teacher might ask the class members to take several minutes, working alone or collaboratively, to list any information they can

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determine about the titration from the pH profile. Sample student responses might include:

• Base is being added to acid.

• The reaction is represented by: HCl + NaOH → NaCl + H₂O • The initial pH = 1.0

• The initial [HCl] = [H+_{] = 10}-pH _{= 10}-1.0 _{= 0.1 mol/L}

• The acid is strong as the initial [H+_{] = the initial [HCl] indicates 100%}

ionization.

• The pH at equivalence = 7.0

• The volume of base required to reach equivalence = 23.5 mL.

• The moles NaOH = moles OH- _{= 25.0 mL x 0.100 mol HCl x 1 mol NaOH =}

1000 mL 1 mol HCl 0.00250 mol NaOH are required to reach equivalence.

• The [NaOH] = 0.00250 mol x 1000 mL = 0.106 mol/L

23.5 mL 1 L

AP Chemistry teachers should reinforce the importance of beginning titration calculations with a balanced chemical equation, ideally in molecular form. This leads to a proper mole ratio that is most likely to be correct when units include the formula of the species they refer to. Any AP Chemistry text will contain a sample curve to which teachers can refer; however, the teacher might prefer to produce a curve using probeware (such as that available from Vernier or Pasco) or to use a variety of simulations available commercially or on the Internet (see suggestions in the laboratory section at the end of this chapter). Production of the curve could be done after, during, or even before the calculations.

Once students have examined a pH profile, or a titration curve for a strong monoprotic acid being titrated with a strong base, the AP Chemistry teacher can discuss the calculation of pH at various points throughout the curve.

Example 1

Calculate the pH of the 25.0 mL sample of 0.100 mol/L solution of HCl being titrated in the profile above.

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43

Strategy:

The student should recognize that pH = - log [H+_{] and that the concentration of a} solution is independent of its volume. Hence, 25.0 mL is irrelevant to the problem at this point. Strong acids such as HCl, ionize completely, hence, [H+_{] = [HCl].}

(AP Chemistry students must be aware of the “BIG 6” strong acids: HCl, HBr, HI, HClO₄, HNO₃, and H₂SO₄ (HClO₃ is included in some texts, although it ionizes less than 100%); H₂SO₄ is only nearly completely ionized for its first ionization) as there is no table of acid ionization constants included with the AP Chemistry Exam.)

Solution:

pH = - log [H+_{] = - log [HCl] = - log (0.100 mol/L) = 1.000}

Example 2

Calculate the pH of the 25.0 mL sample of HCl following the addition of 10.0 mL of 0.106 mol/L NaOH. Note that a “rounder” concentration of base might be used for introduction; however, this is what we’ve determined for our example.

Strategy:

The ICF table is very useful for helping students keep track of everything necessary to solve this problem. If concentrations are to be used (which may be preferable because pH depends upon concentration rather than moles), students must recognize that the concentration of a solution is affected by the addition of water or another solution as both of these cause dilution.

Step one is to dilute. The dilution factors are 25/35ths and 10/35ths,respectively. Step two is to neutralize by applying the ICF table. Step three is to examine the bottom line of the ICF table to determine what affects the pH.

Solution:

[HCl]_in = 0.100 mol/L (25.0 mL) [NaOH]_in = 0.106 mol/L (10.0 mL)

(35.0 mL) (35.0 mL)

= 0.0714 mol/L = 0.0303 mol/L

Neutralization in ICF shows: HCl(aq) + NaOH(aq)→ NaCl(aq) + H₂O(l) I: 0.0714 M 0.0303 M 0 M

C: -0.0303 -0.0303 +0.0303

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44

The bottom line indicates a [HCl] = [H+_{] = 0.0411 mol/L, hence, pH}

= - log [H+_{] = - log (0.0411) = 1.386}

(Note that moles might have been used by multiplying the molarity of each reactant by its volume and using the moles in the “I” line. The changes would then have been 0.00106 moles. The final moles of acid divided by a total volume of 0.0350 L would give the same concentration and consequently, the same pH as shown.)

One final note that should be mentioned is that the ICF table indicates that the [NaOH]_final = 0 mol/L, following the neutralization. This is, of course, impossible, as [H+_][OH-_{] must always = 10}-14 _{= K}

w (at 25oC). Hence, [OH-] = [NaOH] at equivalence is

negligible and approaches 0, though it can never actually equal 0. If [H+_{] = [OH}-_{], then}

each = √ 10-14 _{= 10}-7_{, which is very small compared to the molarity numbers on the}

problem, and so ≈ 0.

Example 3

Calculate the pH of the 25.0 mL sample of HCl following the addition of 23.5 mL of 0.106 mol/L NaOH.

Strategy:

As in example 2, dilute and neutralize, then examine the bottom line of the ICF table. Solution:

[HCl]_in = 0.100 mol/L (25.0 mL) [NaOH]_in = 0.10638 mol/L (23.5 mL)

(48.5 mL) (48.5 mL)

= 0.0515 mol/L = 0.0515 mol/L

Neutralization in ICF shows: HCl(aq) + NaOH(aq)→ NaCl(aq) + H₂O(l) I: 0.0515 M 0.0515 M 0 M-

C: -0.0515 -0.0515 +0.0515

F: 0 0 0.0515

This process shows that the pH of 7 at the equivalence point is due to the formation of 48.5 mL of a 0.0515 mol/L solution of NaCl, which dissociates to form the neutral cation, Na+ _{and an anion, Cl}-_{. As the conjugate base of the strong acid,}

HCl, Cl- _{is too weak to accept any appreciable quantity of protons from the water it is}

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equivalence in the same fashion is helpful as it involves the application of Kw and/or pKw along with the ICF table.

Example 4

Calculate the pH of the 25.0 mL sample of HCl following the addition of 30.0 mL of 0.106 mol/L NaOH solution.

Strategy:

As in example 2, dilute and neutralize, then examine the bottom line of the ICF table. The application of K_w and/or pK_w will be required to convert the [OH-_].

Solution:

[HCl]in = 0.100 mol/L (25.0 mL) [NaOH]in = 0.10638 mol/L (30.0 mL)

(55.0 mL) (55.0 mL)

= 0.0455 mol/L = 0.0580 mol/L

Neutralization in ICF shows: HCl(aq) + NaOH(aq)→ NaCl(aq) + H₂O(l) I: 0.0455 M 0.0580 M 0 M-

C: -0.0455 -0.0455 +0.0455

F: 0 0.0125 0.0455

As explained in example 3, the salt, NaCl, is completely neutral and does not affect the pH at all. However, the [NaOH]_excess is very significant. From this number, it becomes evident that the pOH = - log [OH-_{] = -log[NaOH] = - log (0.0125) = 1.90.}

Consequently, as pK_w = pH + pOH = 14.00 at 25o_{C, the pH = 14.00 – 1.90 = 12.10. The}

alternative approach would be to recognize that [H+_{] = K}

w/[OH-] = 10-14/0.0125 = 8.0 x

10-13 _{M. Hence, pH = - log (8.0 x 10}-13_{) = 12.10.}