# Section 2: Buffered Solutions.  Solutions prepared with common ions have a tendency to resist drastic pH changes even when subjected to the addition.

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Section 2: Buffered Solutions

 Solutions prepared with common ions have a tendency to resist drastic pH changes even when subjected to the addition of strong acids or bases.  Such solutions are called buffered solutions.  Blood is buffered to a pH of about 7.4.  Surface seawater is buffered to a pH between 8.1 and 8.3.

 A buffer resists pH change because it contains …  an acid to neutralize added OH − ions  a base to neutralize added H + ions  But we cannot have the acid and base consume one another through neutralization.

 Therefore, we use weak conjugate acids and bases to prepare our buffers.  For example:  HCH 3 COO and CH 3 COO −  HF and F −  NH 4 + and NH 3  HClO and ClO −

 If we have a solution of HF and NaF, we have HF(aq) and F − (aq) in solution.  If we add a small amount of a strong acid,  we use the F− to neutralize it: F − (aq) + H + (aq) → HF(aq)  If we add a small amount of a strong base,  we use the HF to neutralize it: HF (aq) + OH − (aq) → F − (aq) + H 2 O(l)

 Let’s consider another buffer composed of the weak acid HX and the strong electrolyte MX (M is a metal from Group 1, for example).  The acid dissociation involves both the weak acid and its conjugate base. HX(aq) ⇄ H + (aq) + X − (aq)

 Let’s consider another buffer composed of the weak acid HX and the strong electrolyte MX (M is a metal from Group 1, for example).  The acid dissociation involves both the weak acid and its conjugate base. HX(aq) ⇄ H + (aq) + X − (aq)  The acid dissociation constant is K a = [H + ][X − ] [HX]

 Let’s consider another buffer composed of the weak acid HX and the strong electrolyte MX (M is a metal from Group 1, for example).  The acid dissociation involves both the weak acid and its conjugate base. HX(aq) ⇄ H + (aq) + X − (aq)  Solving for [H + ] [H + ] = K a [HX] [X − ]

[H + ] = K a [HX] [X − ]

[H + ] = K a  The [H + ] and pH depend on two things: [HX] [X − ]

[H + ] = K a  The [H + ] and pH depend on two things:  K a  the ratio of the concentrations of conjugate acid- base pair. [HX] [X − ]

[H + ] = K a  If OH − ions are added, they will react with the acid component to produce water and X −. [HX] [X − ]

[H + ] = K a  If OH − ions are added, they will react with the acid component to produce water and X −. OH − (aq) + HX(aq) → H 2 O(l) + X − (aq) [HX] [X − ]

[H + ] = K a  If OH − ions are added, they will react with the acid component to produce water and X −. OH − (aq) + HX(aq) → H 2 O(l) + X − (aq)  This causes a decrease in [HX] and an increase in [X − ]. [HX] [X − ]

[H + ] = K a  If OH − ions are added, they will react with the acid component to produce water and X −. OH − (aq) + HX(aq) → H 2 O(l) + X − (aq)  This causes a decrease in [HX] and an increase in [X − ].  So long as the amounts of HX and X − are large, the ratio of [HX]/[X − ] does not change. [HX] [X − ]

[H + ] = K a  If H + ions are added, they will react with the base component to produce the acid. [HX] [X − ]

[H + ] = K a  If H + ions are added, they will react with the base component to produce the acid. H + (aq) + X − (aq) → HX (aq) [HX] [X − ]

[H + ] = K a  If H + ions are added, they will react with the base component to produce the acid. H + (aq) + X − (aq) → HX (aq)  This causes a decrease in [X − ] and an increase in [HX]. [HX] [X − ]

[H + ] = K a  If H + ions are added, they will react with the base component to produce the acid. H + (aq) + X − (aq) → HX (aq)  This causes a decrease in [X − ] and an increase in [HX].  So long as the amounts of HX and X − are large, the ratio of [HX]/[X − ] does not change. [HX] [X − ]

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.

 But, there is an alternative method.

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.  But, there is an alternative method.  We start with the expression for finding [H + ] of a buffer.

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.  But, there is an alternative method.  We start with the expression for finding [H + ] of a buffer. [HX] [X − ] [H + ] = K a

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.  But, there is an alternative method.  We then find the negative log of both sides. [HX] [X − ] [H + ] = K a

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.  But, there is an alternative method.  We then find the negative log of both sides. −log[H + ] = −log(K a ) [HX] [X − ]

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.  But, there is an alternative method.  We then find the negative log of both sides. −log[H + ] = −logK a − log [HX] [X − ]

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.  But, there is an alternative method.  We note that −log[H + ] = pH and −logK a = pK a −log[H + ] = −logK a − log [HX] [X − ]

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.  But, there is an alternative method.  We note that −log[H + ] = pH and −logK a = pK a pH = pK a − log [HX] [X − ]

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.  But, there is an alternative method.  We note that − log[HX]/[X − ] = + log[X − ]/[HX] pH= pK a − log [HX] [X − ]

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.  But, there is an alternative method.  We note that − log[HX]/[X − ] = + log[X − ]/[HX] pH= pK a + log [X − ] [HX]

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.  But, there is an alternative method.  We can generalize HX as an acid and X − as a base. pH= pK a + log [X − ] [HX]

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.  But, there is an alternative method.  We can generalize HX as an acid and X − as a base. pH= pK a + log [base] [acid]

 We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.  But, there is an alternative method.  This is called the Henderson-Hasselbach equation. pH= pK a + log [base] [acid]

 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate, NaCH 3 CH(OH)COO? For lactic acid, K a = 1.4 × 10 −4.

 Known:

 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate, NaCH 3 CH(OH)COO? For lactic acid, K a = 1.4 × 10 −4.  Known:K a = 1.4 × 10 −4

 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate, NaCH 3 CH(OH)COO? For lactic acid, K a = 1.4 × 10 −4.  Known:K a = 1.4 × 10 −4 ⇒ pK a = −logK a

 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate, NaCH 3 CH(OH)COO? For lactic acid, K a = 1.4 × 10 −4.  Known:K a = 1.4 × 10 −4 ⇒ pK a = −logK a = 3.85

 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate, NaCH 3 CH(OH)COO? For lactic acid, K a = 1.4 × 10 −4.  Known:K a = 1.4 × 10 −4 ⇒ pK a = −logK a = 3.85 [base] = 0.10 M

 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate, NaCH 3 CH(OH)COO? For lactic acid, K a = 1.4 × 10 −4.  Known:K a = 1.4 × 10 −4 ⇒ pK a = −logK a = 3.85 [base] = 0.10 M [acid] = 0.12 M

 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate, NaCH 3 CH(OH)COO? For lactic acid, K a = 1.4 × 10 −4.  Known:K a = 1.4 × 10 −4 ⇒ pK a = −logK a = 3.85 [base] = 0.10 M [acid] = 0.12 M pH = pK a + log [base] [acid]

 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate, NaCH 3 CH(OH)COO? For lactic acid, K a = 1.4 × 10 −4.  Known:K a = 1.4 × 10 −4 ⇒ pK a = −logK a = 3.85 [base] = 0.10 M [acid] = 0.12 M pH = 3.85+ log 0.10 0.12

 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate, NaCH 3 CH(OH)COO? For lactic acid, K a = 1.4 × 10 −4.  Known:K a = 1.4 × 10 −4 ⇒ pK a = −logK a = 3.85 [base] = 0.10 M [acid] = 0.12 M pH = 3.85+ log = 3.85 + log(0.83) 0.10 0.12

 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate, NaCH 3 CH(OH)COO? For lactic acid, K a = 1.4 × 10 −4.  Known:K a = 1.4 × 10 −4 ⇒ pK a = −logK a = 3.85 [base] = 0.10 M [acid] = 0.12 M pH = 3.85+ log = 3.85 − 0.08 0.10 0.12

 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate, NaCH 3 CH(OH)COO? For lactic acid, K a = 1.4 × 10 −4.  Known:K a = 1.4 × 10 −4 ⇒ pK a = −logK a = 3.85 [base] = 0.10 M [acid] = 0.12 M pH = 3.85+ log = 3.85 − 0.08 = 3.77 0.10 0.12

 Use Sample Exercise 17.3 to help solve homework exercises 17.21 through 17.24 (pages 759 − 760).

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)

 Known:

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9 [base] = 0.10 M

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9 [base] = 0.10 M [H + ] = K a [acid] [base]

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9 [base] = 0.10 M [H + ] = K a ⇒ [acid] = [acid] [base] [H + ][base] KaKa

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9 [base] = 0.10 M [acid] = [H + ][base] KaKa

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9 [base] = 0.10 M [acid] = = [H + ][base] KaKa (1.0 × 10 −9 )(0.10) 5.6 × 10 −10

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9 [base] = 0.10 M [acid] = = 0.18 M [H + ][base] KaKa

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9 [base] = 0.10 M [acid] = = 0.18 M = [NH 4 + ] [H + ][base] KaKa

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9 [base] = 0.10 M [NH 4 + ] = 0.18 M

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9 [base] = 0.10 M [NH 4 + ] = 0.18 M ⇒

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9 [base] = 0.10 M [NH 4 + ] = 0.18 M ⇒ n = M × V

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9 [base] = 0.10 M [NH 4 + ] = 0.18 M ⇒ n = (0.18 M)(2.0 L)

 How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH 4 Cl.)  Known:K b = 1.8 × 10 −5 ⇒ K a = 5.6 × 10 −10 pH = 9.00 ⇒ [H + ] = 1.0 × 10 −9 [base] = 0.10 M [NH 4 + ] = 0.18 M ⇒ n = 0.36 moles

 Use Sample Exercise 17.4 to help solve homework exercises 17.25 through 17.26 (page 760).

 There are two very important characteristics of buffers.  the buffer capacity  the effective pH range

 The buffer capacity is the amount of acid or base a buffer can neutralize before the pH begins to change to an appreciable degree.  Buffer capacity depends on …  the amount of buffer materials  the identity of the buffer materials

 The pH range of any buffer is the range of pH over which a buffer works to effectively neutralize an acid or a base.  The optimal situation is where the concentration of the buffer acid is equal to the concentration of the buffer base.  In that case, pH = pK a.  Buffers usually have a pH range of about ±1 pH unit of pK a.

 Next, we’ll look at what happens when we add a strong acid or strong base to a buffer.  This is going to be a quantitative approach (that means numbers!).  Remember … ▪ Strong acids react with weak bases to completion. ▪ H + + ClO − → HClO ▪ Strong bases react with weak acids to completion. ▪ OH − + CH 3 COOH → H 2 O + CH 3 COO −

 To calculate how the pH of a buffer changes when adding a strong acid or base, we follow the following strategy.  First, consider the acid-base neutralization reaction and determine its effect on [HX] and [X − ].  Use the K a and new values for [HX] and [X − ] to calculate [H + ]. ▪ Use i-c-e table or Henderson-Hasselbach (easiest).

 A buffer is made by adding 0.300 mol CH 3 COOH and 0.300 mol CH 3 COONa to enough water to make 1.00 L of solution.

 A buffer is made by adding 0.300 mol CH 3 COOH and 0.300 mol CH 3 COONa to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1).

a) Calculate the pH of this solution after 0.020 mol NaOH is added.

 A buffer is made by adding 0.300 mol CH 3 COOH and 0.300 mol CH 3 COONa to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). a) Calculate the pH of this solution after 0.020 mol NaOH is added. b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization.

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. OH − + CH 3 COOH → CH 3 COO − + H 2 O

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. OH − + CH 3 COOH → CH 3 COO − + H 2 O n initial 0.0200.300

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. OH − + CH 3 COOH → CH 3 COO − + H 2 O n initial 0.0200.300 n chg −0.020 +0.020

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. OH − + CH 3 COOH → CH 3 COO − + H 2 O n initial 0.0200.300 n chg −0.020 +0.020 n final 0.0000.2800.320

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. OH − + CH 3 COOH → CH 3 COO − + H 2 O This give us:[CH 3 COOH] = 0.280 M n initial 0.0200.300 n chg −0.020 +0.020 n final 0.0000.2800.320

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. OH − + CH 3 COOH → CH 3 COO − + H 2 O This give us:[CH 3 COOH] = 0.280 M [CH 3 COO − ] = 0.320 M n initial 0.0200.300 n chg −0.020 +0.020 n final 0.0000.2800.320

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. OH − + CH 3 COOH → CH 3 COO − + H 2 O This give us:[CH 3 COOH] = 0.280 M [CH 3 COO − ] = 0.320 M

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. This give us:[CH 3 COOH] = 0.280 M [CH 3 COO − ] = 0.320 M

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. This give us:[CH 3 COOH] = 0.280 M [CH 3 COO − ] = 0.320 M

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. This give us:[CH 3 COOH] = 0.280 M [CH 3 COO − ] = 0.320 M  Using Henderson-Hasselbach:

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. This give us:[CH 3 COOH] = 0.280 M [CH 3 COO − ] = 0.320 M  Using Henderson-Hasselbach: pH = pK a + log [base]

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. This give us:[CH 3 COOH] = 0.280 M [CH 3 COO − ] = 0.320 M  Using Henderson-Hasselbach: pH = 4.74+ log [0.280] [0.320]

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. This give us:[CH 3 COOH] = 0.280 M [CH 3 COO − ] = 0.320 M  Using Henderson-Hasselbach: pH = 4.74+ log(1.14)

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. This give us:[CH 3 COOH] = 0.280 M [CH 3 COO − ] = 0.320 M  Using Henderson-Hasselbach: pH = 4.74+ log(1.14) = 4.74 + 0.06

a) Calculate the pH of this solution after 0.020 mol NaOH is added.  First we need to do the neutralization. This give us:[CH 3 COOH] = 0.280 M [CH 3 COO − ] = 0.320 M  Using Henderson-Hasselbach: pH = 4.74+ log(1.14) = 4.74 + 0.06 = 4.80

b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.

In pure water, NaOH is fully dissociated.

b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water. In pure water, NaOH is fully dissociated. Therefore, [OH − ] = 0.020 mol/1.00 L

b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water. In pure water, NaOH is fully dissociated. Therefore, [OH − ] = 0.020 mol/1.00 L = 0.020 M.

b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water. In pure water, NaOH is fully dissociated. Therefore, [OH − ] = 0.020 mol/1.00 L = 0.020 M. pOH = −log[OH − ]

b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water. In pure water, NaOH is fully dissociated. Therefore, [OH − ] = 0.020 mol/1.00 L = 0.020 M. pOH = −log[OH − ] = −log(0.020)

b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water. In pure water, NaOH is fully dissociated. Therefore, [OH − ] = 0.020 mol/1.00 L = 0.020 M. pOH = −log[OH − ] = −log(0.020) = 1.70

b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water. In pure water, NaOH is fully dissociated. Therefore, [OH − ] = 0.020 mol/1.00 L = 0.020 M. pOH = −log[OH − ] = −log(0.020) = 1.70 pH = 14.00 − pOH

b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water. In pure water, NaOH is fully dissociated. Therefore, [OH − ] = 0.020 mol/1.00 L = 0.020 M. pOH = −log[OH − ] = −log(0.020) = 1.70 pH = 14.00 − pOH = 14.00 − 1.70

b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water. In pure water, NaOH is fully dissociated. Therefore, [OH − ] = 0.020 mol/1.00 L = 0.020 M. pOH = −log[OH − ] = −log(0.020) = 1.70 pH = 14.00 − pOH = 14.00 − 1.70 = 12.30

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