5.4
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5.4 pH and pOH Warm Up 1. Complete the following table. [H3O+]
Solution
[OH−]
A. 1.0 M NaOH B. 1.0 M HCl 2. Determine the [H3O+] and [OH−] in the solution formed when equal volumes of solution A and solution B are combined.
3. Calculate the volume of solution A required to obtain 1.0 mol of H3O+ ions.
The pH Scale
Look at your answers to question 1 in the Warm Up and consider the magnitude of the difference between the [H3O+] and [OH−] in solution A and solution B — it is 100 trillion times! Consider also that during the neutralization described in question 2, each ion concentration changes instantly by 10 million times to equal the other. Now think about the volume your answer to question 3 actually represents: 1014 L of 1.0 M NaOH solution would be required to obtain 1.0 mol of hydronium ions! Let’s try to appreciate how large a volume this is. The average flow rate of large river is approximately 3.5 × 106 L/s (Figure 5.4.1). At this rate, to observe 1014 L of water flow by, you would have to watch this mighty river for more than 330 days! The point behind the above discussion is that a linear arithmetic scale is not only inconvenient, it is cumbersome. It is also often inadequate to use when representing the typically very small hydronium and hydroxide concentrations in most aqueous solutions and the extent to which they can change in a neutralization reaction. A much more convenient and compact approach was introduced in 1909 by the great Danish chemist S.P. Sorensen. Sorensen’s scale is called the pH (“potency” or “power” of hydrogen) scale. It is a logarithmic or “power of 10” way to specify the concentration of hydronium ions in a solution. We define pH as the negative logarithm of the molar concentration of hydronium ions: pH = –log [H3O+]
Figure 5.4.1 A powerful river
flows through a narrow valley.
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The logarithm of a number is the power to which 10 must be raised to obtain that number. For example: log 100 = log (1 × 102) = 2.0 log 0.01 = log (1 × 10–2) = –2.0
Logarithms and Significant Figures
When we take the logarithm of a number, we must be careful to record the answer to the proper number of significant figures. To understand how to do this, consider the following example:
Determine the pH of a 0.0035 M H3O+ solution and record the answer to the proper number of significant figures.
Solution: Note that in the molar concentration, there are two significant figures because the leading zeros are not considered significant. It should make sense to you that any operation done on those zeros yields a value that is also not considered significant. In the solution below, the significant figures are underlined:
pH = –log [H3O+] = –log (0.0035) (two significant figures) = –log (3.5 × 10–3)
(Note that the log of the product of two numbers equals the sum of the logs of those numbers so that: log (A × B) = log A + log B)
= –log 3.5 (significant) + –log 10–3
(The exponent 3 is not a measured value. Rather, it is an exact number and consequently does not limit the number of significant figures in the answer.)
= –0.5441 (significant) + 3 = 2. 4559 = 2.46 (two significant figures)
We can generalize this into a rule governing logarithms and significant figures: When taking the logarithm of a number, the result must have the same number of decimal places as there are significant figures in the original number.
Measuring pH
The pH of a solution is usually determined with an acid-base indicator or, more precisely, by using an instrument called a pH meter (Figure 5.4.2). (Acid-base indicators will be discussed in more detail in Chapter 6.) A pH meter consists of specially designed electrodes that are sensitive to the concentration of H3O+ ions. When the electrodes are dipped into a solution, a voltage between the measuring electrode and a reference electrode is generated that depends on the solution’s pH. That voltage is read on a meter which is calibrated in pH units.
Figure 5.4.2 A pH meter
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Quick Check 1. Convert each of the following hydronium concentrations to pH values. Make sure to record your answers to the proper number of significant figures. Solution
[H3O+]
orange juice
3.2 × 10–4 M
milk of magnesia
2.52 × 10–11 M
stomach acid
0.031 M
pH
2. During a titration, if the pH of a solution decreased quickly by 5 units from 8 to 3 at the equivalence point, did the [H3O+] increase or decrease and by how much? __________________________________________________________________________________________ 3. A 100.0 mL sample of an aqueous solution labeled as 0.10 M H3O+ is diluted to 1.0 L. Has the pH of the solution increased or decreased and by how much? __________________________________________________________________________________________
Antilogarithms
You will be expected to determine the pH of a solution given its hydronium concentration. You must also be able to convert pH values into hydronium concentrations in the reverse process. This is accomplished using the following relationship: [H3O+] = 10–pH
This is sometimes called taking the antilogarithm, and in this case, we are taking the antilogarithm of the negative pH value. It is simply expressing the negative logarithm in its original exponential form. Depending on the type of calculator you use, you should become familiar with the calculation steps to perform this operation. Usually, this requires employing the “10x ,” “ Yx ,” or “inverse” “log” keys.
Sample Problem 5.4.1 — Converting pH to [H3O+]
The pH of a beaker of lemon juice is measured and found to be 2.31. Calculate the concentration of hydronium ions in this solution.
What to Think About
How to Do It
1. Because the pH is well below 7.00, expect the hydronium concentration to be much greater than 1.0 × 10–7 M. The solution is therefore acidic.
[H3O+] = 10–pH [H3O+] = 10–2.31
2. Two decimal places in the pH value will correspond to a molar concentration recorded to two significant figures.
[H3O+] = 4.9 × 10–3 M
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Practice Problems 5.4.1 — Converting pH to [H3O+] 1. Complete the following table and record your answers to the proper number of significant figures: [H3O+]
No. of sig. figures
pH
5.00 × 10–5 M 3.34 6.4 × 10–11 M 6.055 0.00345 M 2. During a titration, the pH at the equivalence point changed quickly from 4.35 to 9.65. Calculate the [H3O+] just before and just after the equivalence point was reached.
3. A student calculates the pH of a sample of concentrated nitric acid to be –1.20. She thinks she has made a mistake because the value is negative. Calculate the hydronium concentration in this acid solution and decide if an error was made. Why do you think expressing pH values for concentrated strong acid solutions are not really necessary?
Expanding the “p” Concept
The success of the pH concept for representing small hydronium concentrations allowed it to be extended to cover other typically small values. In general: pX = –log X Accordingly, we can also represent the hydroxide concentrations in aqueous solutions in a logarithmic way and call that value pOH. pOH = –log [OH−] For example, in a 0.10 M solution of NaOH, we know that the [OH−] = 0.10 M. This allows us to determine the pOH as follows:
pOH = –log 0.10 = –log (1.0 × 10–1) = 1.00
Recall that the concentrations of hydronium and hydroxide ions in aqueous solutions are inversely related through the equilibrium constant we call Kw. This always equals 1.00 × 10–14 at 25°C as given by: [H3O+] [OH−] = Kw = 1.00 × 10–14 © Edvantage Interactive 2018
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We can use this to derive another very useful relationship for any aqueous solution. Begin with: [H3O+][OH−] = Kw Take the negative log of both sides:
[
]
–log [H3O+][OH−] = –log Kw This is simply:
= –log [H3O+] + –log [OH−] = –log Kw
pH + pOH = pKw The above relationship is always true for any aqueous solution at any temperature. If we assume a temperature of 25°C, then: pH + pOH = 14.00 In Figure 5.4.3(a), note the graphical relationship showing the inverse relationship between hydronium and hydroxide concentration in aqueous solutions as given by [H3O+][OH−] = 10–14. Note how the negative logarithm of both sides of the relationship translates to the graphical relationship in (b) as given by pH + pOH = 14.00. OH– Concentration vs. H3O+ Concentration
pH vs. pOH
14 12
basic (pH >7, pOH <7
pH (-log [H3O+])
10
[OH–]
8 neutral
(pH = pOH = 7
6 4 acidic (pH <7, pOH >7 2 0
(a)
[H3O+]
0
2
4
6
8
10
12
14
pOH (–log [OH–])
(b)
Figure 5.4.3 (a) The inverse relationship between hydronium and hydroxide concentration in aqueous solutions; (b) This graph uses the
negative logs of both sides of the relationship shown in (a).
This relationship allows us to add to our criteria for designating aqueous solutions as acidic, basic, or neutral by incorporating both pH and pOH (Table 5.4.1). Table 5.4.1 pH and pOH Criteria for Designating Aqueous Solutions as Acidic, Basic, or Neutral
Aqueous Solution at Any Temperature
Aqueous Solution at 25°C
Result
pH = pOH
pH = pOH = 7
Solution is neutral.
pH < pOH
pH < 7 and pOH > 7
Solution is acidic.
pH > pOH
pH > 7 and pOH < 7
Solution is basic.
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This relationship also means that for any aqueous solution, if we are given any one of pH, pOH, [H3O+], or [OH−], we can always calculate the remaining three. Consider Figure 5.4.4. Kw/[H3O+]
[H3O+]
[OH–]
[H3O+]•[OH–] = Kw Kw/[OH–]
–log [OH–]
–log [H3O+] 10-pOH
10–pH 14.0 – pH
pOH
pH + pOH = pKw
pH
14.0 – pOH
Figure 5.4.4 This diagram shows how, by knowing one of pH, pOH, [H3O+], or [OH−], we can calculate the other three.
We can therefore summarize the relationship between pH, pOH, [H3O+], and [OH−] for any aqueous solution at 25°C as shown in Figure 5.4.5. On the chart, consider the “cursor” that sits over the middle of the scale, highlighting a neutral solution. Imagine that this cursor is transparent and movable. It could be moved to the left or the right to identify the conditions associated with increasingly acidic or basic solutions respectively. Solution becomes more acidic.
Solution becomes more basic.
[H3O+]
1
10–1
10–2
10–3
10–4
10–5
10–6
10–7
10–8
10–9
10–10
10–11
10–12
10–13
10–14
pH
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
pOH
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
[OH−]
10–14
10–13
10–12
10–11
10–10
10–9
10–8
10–7
10–6
10–5
10–4
10–3
10–2
10–1
1
CURSOR
Solution is
neutral.
Figure 5.4.5 The “cursor” on the chart highlights the pH, pOH, [H3O+], and [OH−] for any aqueous solution at 25°C.
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Sample Problem 5.4.2 — Calculating pH, pOH, [H3O+], and [OH−]
A 0.10 M aqueous solution of Aspirin (acetylsalicylic acid) at 25°C is found to have a pH of 2.27. Determine the [H3O+], [OH−], and pOH for this solution.
What to Think About
How to Do It
1. This solution is acidic so expect the [H3O+] to be greater than [OH−]. It is very important to recognize which is the predominant ion in a solution such as this. For acids, the [H3O+] always predominates. This means it is more prevalent.
2. Two decimal places in the pH correspond to two significant figures in the concentration. Be sure to keep all figures through the entire calculation. Only round to the appropriate number of significant figures at the very end. 3. Consider Figure 5.4.4 and note that there is more than one order in which the calculations can be performed. For example, we could answer the questions in the order they are asked, or we could begin by calculating pOH and then determine [OH−] and [H3O+].
Perform the calculations in the order they are asked. In terms of Figure 5.4.3, begin in the lower left corner and move around clockwise to answer the questions: [H3O+] = 10–pH = 10–2.27 = 5.4 × 10–3 M [OH−] = Kw /[H3O+] –14 [OH−] = 1.0 × 10 –3 = 1.9 × 10–12 M 5.4 × 10 pOH = –log (1.86 × 10–12 M) = 11.73 OR pOH = 14.00 – 2.27 = 11.73
[OH−] = 10–11.73 = 1.9 × 10–12 M –14 [H3O+] = 1.0 × 10–12 = 5.4 × 10–3 M 1.9 × 10
Practice Problems 5.4.2 — pH and pOH (Unless otherwise indicated, assume all aqueous solutions are at 25°C.) 1. Complete the following table. Solution
[H3O+]
[OH−]
pH
Orange Juice Tears
pOH
Acidic/Basic/Neutral?
10.5 3.98 × 10–8 M
Blood Milk
7.40 3.16 × 10–8 M
2. Consider the equations above relating pH, pOH, [H3O+], and [OH−] for an aqueous solution, and complete the following statements: (a) As a solution becomes more acidic, both [H3O+] and pOH ___________ (increase or decrease) and both [OH−] and pH ___________ (increase or decrease). (b) A basic solution has a pOH value that is ___________ (greater or less) than 7, and [H3O+] that is ___________ (greater or less) than 10–7 M. (c) If the pH of a solution equals 14.0, the [OH−] equals ________ M. (d) If the pOH of a solution decreases by 5, then the [H3O+] has ___________ (increased or decreased) by a factor of ___________. 3. For pure water at 10°C, Kw = 2.55 × 10–15. Calculate the pH of water at 10°C and state whether the water is acidic, basic, or neutral.
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When Strong Acids and Strong Bases Are Mixed
Look again at your answer to the Warm Up question 2 at the beginning of this section. In this example, hydronium ions from a strong acid reacted with an equal number of moles of hydroxide ions from a strong base. The resulting aqueous solution was neutral because neither ion was in excess following the neutralization reaction. Therefore, the [H3O+] and [OH−] were both 1.0 × 10–7 M, so both the pH and pOH were equal to 7.00. However, if an excess of either hydronium or hydroxide ions remains in a solution following the reaction of a strong acid with a strong base, the solution will not be neutral. Recall that the net ionic equation for the reaction of a strong acid with a strong base is given by: H3O+(aq) + OH−(aq) → 2 H2O(l) This allows us to easily determine which ion will be in excess and by how much after a strong acid and a strong base are mixed together. We can then determine pH and/or pOH values. A convenient approach is to calculate the starting or diluted [H3O+] and [OH−] when the solutions are poured together but before they react and then to simply subtract the lesser concentration from the greater concentration to determine the final concentration of the ion in excess. Let us designate those starting concentrations as: [H3O+]ST and [OH−]ST. Depending on which of these two concentrations is greater, we use one of the two equations below to calculate the pH or pOH: If [H3O+]ST > [OH−]ST, then we use: [H3O+]XS = [H3O+]ST – [OH−]ST to calculate pH or pOH. If [OH−]ST > [H3O+]ST , then we use: [OH−]XS = [OH−]ST – [H3O+]ST to calculate pH or pOH.
Sample Problem 5.4.3(a) — Mixing Strong Acids and Bases Calculate the pH of the solution resulting from mixing 30.0 mL of 0.40 M HNO3 with 70.0 mL of 0.20 M NaOH.
What to Think About 1. Nitric acid is a strong acid and so the concentration of the acid is also the [H3O+] in that solution due to 100% ionization. 2. The final volume will be 100.0 mL so each ion concentration will be reduced accordingly upon mixing.
How to Do It
[H3O+]ST = 0.40 M × 30.0 mL = 0.12 M 100.0 mL [OH−]ST = 0.20 M × 70.0 mL = 0.14 M 100.0 mL [OH−]XS = 0.14 M – 0.12 M = 0.02 M pOH = –log (0.02) = 1.7 And so pH = 14.0 – 1.7 = 12.3
An alternative approach would be to calculate the total number of moles of each ion and then subtract the lesser from the greater. Then divide the difference by the final volume to obtain the concentration of the ion is excess.
mol H3O+ mol = 0.0300 L × 0.40 = 0.012 mol H3O+ L – mol OH− = 0.0700 L × 0.20 mol OH = 0.014 mol OH− L − mol OH in excess = 0.014 mol – 0.012 mol = 0.002 mol OH− in excess 0.002 mol OH– The final [OH−] = = 0.02 M 0.1000 L pOH = –log 0.02 = 1.7 so pH = 14.0 – 1.7 = 12.3 H3O+
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Sample Problem 5.4.3(b) — Mixing Strong Acids and Bases Determine the pH of the solution that results when 50.0 mL of 0.200 M H2SO4 is mixed with 100.0 mL of 0.400 M NaOH.
What to Think About 1. The final volume will be 150.0 mL and so each ion concentration should be diluted accordingly. 2. H2SO4 is a diprotic acid and so the [H3O+]ST should be calculated accordingly. 3. The final pH must be recorded to three decimal places.
How to Do It
[H3O+]ST
2 mol H3O+ 50.0 mL mol H2SO4 = 0.200 × × mol H2SO4 150.0 mL L = 0.1333 M
[OH−]ST = 0.400 M × 100.0 mL = 0.2667 M 150.0 mL − [OH ]XS = 0.2667 M – 0.1333 M = 0.1333 M
pOH = –log (0.1333) = 0.875
pH = 14.000 – 0.875 = 13.125
Sample Problem 5.4.3(c) — Mixing Strong Acids and Bases A student adds 35.0 mL of an HCl solution with a pH of 2.00 to 15.0 mL of NaOH solution with a pH of 12.00. Calculate the pH of the final solution.
What to Think A bout
How to Do It
1. The predominant ion in the first solution (with a pH of 2.00) is clearly the hydronium ion, and the predominant ion in the second (pH = 12.00) is hydroxide.
In the basic solution: pOH = 14.00 – 12.00 = 2.00
2. Accordingly, determine each ion concentration as shown previously. 3. Record two decimal places in the final pH.
In the acidic solution: [H3O+] = 10–2.00 = 0.010 M [OH−] = 10–2.00 = 0.010 M [H3O+]ST = 0.010 M × 35.0 mL = 0.0070 M 50.0 mL [OH−]ST = 0.010 M × 15.0 mL = 0.0030 M 50.0 mL + [H3O ]XS = 0.0070 M – 0030 M = 0.0040 M
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pH = –log (0.0040) = 2.40
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Sample Problem 5.4.3(d) —Mixing Strong Acids and Bases What mass of NaOH must be added to 500.0 mL of a solution of 0.020 M HI to obtain a solution with a pH of 2.50?
What to think about 1. The final solution still has an excess of H3O+ ions so the equation that applies is: [H3O+]XS = [H3O+]ST – [OH−]ST 2. Solve for [OH−]ST because both the [H3O+]XS and [H3O+]ST are provided either directly or indirectly in the question. Therefore manipulate the above equation to give: [OH−]ST = [H3O+]ST – [H3O+]XS
Then convert your answer from moles per liter to grams required for 0.500 L.
How to Do It
[H3O+]ST = 0.020 M [H3O+]XS = 10–2.50 = 0.00316 M [OH−]ST = [H3O+]ST – [H3O+]XS
= 0.020 M – 0.00316 M
= 0.0168 M NaOH
0.0168 mol × 40.0 g NaOH × 0.500 L L mol = 0.34 g NaOH (2 significant figures)
Practice Problems 5.4.3 — Mixing Strong Acids and Bases 1. Calculate the pH of the solution that results from mixing 25.0 mL of 0.40 M HCl with 15.0 mL of 0.30 M KOH.
2. A solution of HCl has a pH of 0.60. A 1.0 g sample of NaOH is dissolved in 250.0 mL of this HCl solution. Calculate the pOH of the resulting solution, assuming no volume change.
3. Calculate the pH of the solution that results when 25.0 mL of a solution with a pOH of 12.00 is mixed with 45.0 mL of a solution with a pH of 11.00.
4. What mass of HCl(g) must be dissolved in 1.50 L of a NaOH solution having a pH of 11.176 to produce a solution with a pH of 10.750? (Assume no volume change.)
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