Buffer Effectiveness 19

Buffer Effectiveness 19

Buffer Effectiveness What makes a buffer effective? A buffer should be able to neutralize small to moderate amounts of added acid or base Too much added acid or base will destroy a buffer What influences the effectiveness of a buffer? The relative amounts of acid/base and its conjugate The absolute concentrations of acid/base and its conjugate

Buffer Effectiveness Let s consider a buffer solution that s made from an acid and its conjugate base This buffer is most effective (resistant to ph changes) when: 1 [acid] and [conjugate base] are almost equal The accepted range is 0.10 [base]/[acid] 10 2 The concentrations are both are high (the buffer is concentrated, not dilute)

Buffer Range 22

Buffer Range Again, let s consider a buffer solution that s made from an acid and its conjugate base Since 0.10 [base]/[acid] 10, we can use the Henderson-Hasselbach equation to determine an appropriate range of ph values for an effective buffer ph = pk a + log (0.10) and ph = pk a + log (10) So our range is: pk a -1 < ph < pk a + 1

Buffer Range pk a -1 < ph < pk a + 1 What does this mean? A weak acid with a pk a of 5.0 is used to prepare a buffer in the range of 4.0 6.0. You need to calculate and adjust the relative amounts of [conjugate base] and [acid] to get an ideal ratio. The closer to ph of 5.0, the more effective the buffer will be.

Examples: Preparing a Buffer 25

Preparing a Buffer Look at the acids below. Which would you choose to combine with its sodium salt to make a solution buffered at ph 4.25? For the best choice, calculate the ratio of the conjugate base to the acid required to attain the desired ph. a. Chlorous acid (HClO 2 ) pk a = 1.95 b. Nitrous acid (HNO 2 ) pk a = 3.34 c. Formic acid (HCHO 2 ) pk a = 3.74 d. Hypochlorous acid (HClO) pk a = 7.54

Preparing a Buffer Formic acid is the best choice because the pk a is the closest to the ph value a. Chlorous acid (HClO 2 ) pk a = 1.95 b. Nitrous acid (HNO 2 ) pk a = 3.34 c. Formic acid (HCHO 2 ) pk a = 3.74 d. Hypochlorous acid (HClO) pk a = 7.54

Preparing a Buffer What is the desired ratio? Formic acid (HCHO 2 ) pk a = 3.74 Use your equation ph = pk a + log [base]/[acid] and plug-in values for ph and pk a. So 4.25 = 3.74 + log [base]/[acid] 0.51 = log [base]/[acid] 10 0.51 = [base]/[acid] 3.24 = [base]/[acid]

Try This: A student is using an acetic acid/sodium acetate buffer solution. Explain how they should determine the ratio of acid to conjugate base needed to maintain a ph of 5.00.

Try This: Look at the acids below. Which would you choose to combine with its sodium salt to make a solution buffered at ph 7.35? If you have 500.0 ml of a 0.10 M solution of the acid, what mass of the corresponding sodium salt would you need to make the buffer? a. Chlorous acid (HClO 2 ) pk a = 1.95 b. Nitrous acid (HNO 2 ) pk a = 3.34 c. Formic acid (HCHO 2 ) pk a = 3.74 d. Hypochlorous acid (HClO) pk a = 7.54

Buffer Capacity 31

Buffer Capacity Buffer Capacity = the amount of acid or base that we can add to a buffer without causing a large change in ph. Buffer capacity increases 1 With increasing absolute concentrations of the buffer components 2 As the relative concentrations of the buffer components become more similar to each other (the ratio approaches 1)

Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer

Think A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? a. Adding 0.050 mol HCl b. Adding 0.050 mol of NaOH c. Adding 0.050 mol of NaF d. None of the above

Sample Question Which of the following will not produce a buffered solution? a. 100 ml of 0.1 M Na 2 CO 3 and 50 ml of 0.1 M HCl b. 100 ml of 0.1 M NaHCO 3 and 25 ml of 0.2 M HCl c. 100 ml of 0.1 M Na 2 CO 3 and 75 ml of 0.2 M HCl d. 50 ml of 0.2 M Na 2 CO 3 and 5 ml of 1.0 M HCl e. 100 ml of 0.1 M Na 2 CO 3 and 50 ml of 0.1 M NaOH

Sample Question You are given a solution of the weak base Novocain, Nvc (ph is 11.00) You add to this solution a small amount of a salt containing the conjugate acid of Novocain, NvcH + Which of the following statements is true? a. Both the ph and the poh increase b. Both the ph and the poh decrease c. Both the ph and the poh remain unchanged d. ph increases and the poh decreases e. ph decreases and the poh increases

Sample Question Consider a buffered solution at ph of 4.00 made with HF (K a = 7.2 10-4 ) and NaF Which of the following statements is true? All of the concentrations below refer to equilibrium concentrations a. [HF] = [F ] b. [HF] = [H + ] c. [HF] > [F ] d. [H + ] = [F ] e. [HF] < [F ]

Sample Question A solution will have the most effective buffering capacity when the concentrations of the buffering components are: a. Small b. Large c. Concentration does not matter

Titrations

Titrations Recall: a type of volumetric analysis we can use to determine the amount of a certain substance is a titration. Often, we look at an acid-base neutralization rxn In a titration, a standard solution (of KNOWN concentration) is added gradually to a solution of unknown concentration until the chemical reaction is complete. Since we know the concentration and the volume added of the standard solution, as well as the volume of the unknown solution, we can calculate the concentration of the unknown solution.

Titrations Equipment Required: Beaker/flask Measuring pipette or buret ph meter and/or indicator

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Terms to know: Titrations Titrant = standard solution (known M) Analyte = substance being analyzed (unknown M) Equivalence point = When moles of OH - = moles of H 3 O + (neutralization has occurred) Indicator = Substance added that will undergo a color change near the equivalence point End point = When the indicator changes the color of the solution. This may not match the equivalence point, depending on the range of ph values where the indicator changes color Standard Solution = solution of known concentration that is slowly added to solution of unknown conc

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Titrations Acid-Base Titration Method Using Indicator: 1. Analyte solution (of unknown M) is placed in a flask or beaker 2. A small amount of indicator is added 3. Titrant is placed in a burette and slowly added to the analyte and indicator mixture 4. The process is stopped when the indicator causes a change in the color of the solution 5. The change in volume is used to determine the volume of the analyte solution

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Titrations Note: Some titrations require boiling to get rid of CO 2 produced, which will form carbonic acid, buffer the solution, and lead to inaccurate data

Titrations A titration curve is a plot of the ph of a solution during a titration It is typically the volume added (our standard solution) vs. ph. The shape of the curve depends on the acid/base being used (in particular, depending on the acid/ base strength) as well as whether or not the acid or base is our titrant.

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Titration of a Strong Acid with a Strong Base

Strong Acid w/ Strong Base Let s look at how we can plot the titration curve for the titration of 25.0 ml of 0.100 M HCl with 0.100 M NaOH. We will need to determine the volume required to reach the equivalence point We will need to determine the ph at various volumes before, during, and after the equivalence point Keep in mind: I am adding NaOH to HCl BEFORE equivalence there is excess H 3 O + AFTER equivalence there is excess OH -

Strong Acid w/ Strong Base For a strong acid and strong base: BEFORE equivalence - since H 3 O + is in excess, you calculate the number of moles of added OH - from the initial moles of H 3 O +, and then divide by the total volume AT equivalence H 3 O + and OH - are completely neutralized. Only water contributes to [H 3 O + ], so the ph = 7. AFTER equivalence since OH - is in excess, calculate [OH - ] by subtracting initial moles of H 3 O + from the number of moles of added OH -, then divide by the total volume. Determine ph from poh or determine [H 3 O + ] from [OH - ] to determine ph.

Strong Acid w/ Strong Base Let s look at how we can plot the titration curve for the titration of 25.0 ml of 0.100 M HCl with 0.100 M NaOH. Let s find the initial ph value For the titration of a strong acid and strong base, the ph initial = ph strong acid [HCl] = [H 3 O + ] = 0.100 M So ph = -log(0.100 M) ph =1.000

Strong Acid w/ Strong Base Let s look at how we can plot the titration curve for the titration of 25.0 ml of 0.100 M HCl with 0.100 M NaOH. We want moles acid à moles of base 25.0 ml = 0.0250 L 0.0250 L (0.100 mol/l HCl) = 0.00250 mol HCl 0.00250 mol HCl = 0.00250 mol OH - at equivalence 0.00250 mol OH - (1 L/ 0.100 mol NaOH) = 0.0250 L NaOH The equivalence point will be reached when we add 25 ml of NaOH That means the ph will be 7

Our Data Table So Far ml NaOH added ph 0.00 ml 1.00 5.00 ml 10.00 ml 15.00 ml 20.0 ml 25.0 ml 7.00 30.0 ml 35.0 ml 40.0 ml 50.0 ml

Strong Acid w/ Strong Base Let s look at how we can plot the titration curve for the titration of 25.0 ml of 0.100 M HCl with 0.100 M NaOH. Before equivalence is when we add less than 25 ml NaOH calculate the number of moles of added OH - from the initial moles of H 3 O +, and then divide by the total volume Let s look at the ph after adding 5.00 ml of NaOH

Strong Acid w/ Strong Base Let s look at how we can plot the titration curve for the titration of 25.0 ml of 0.100 M HCl with 0.100 M NaOH. Find the ph after adding 5.00 ml of NaOH 0.00500 L NaOH (0.100 mol/l) = 0.000500 mol NaOH OH - H 3 O + Initial (moles) 0.00 mol 0.00250 mol Addition 0.000500 mol -- After addition 0 (all rxts w/ H 3 O + ) 0.00200 mol

Strong Acid w/ Strong Base Let s look at how we can plot the titration curve for the titration of 25.0 ml of 0.100 M HCl with 0.100 M NaOH. Now we divide the moles of H 3 O + by the TOTAL volume So [H 3 O + ] = 0.00200 mol H 3 O + / (0.0250 L initial + 0.00500 L added) [H 3 O + ] = 0.0667 M and ph = 1.18 OH - H 3 O + Initial (moles) 0.00 mol 0.00250 mol Addition 0.000500 mol -- After addition 0 (all rxts w/ H 3 O + ) 0.00200 mol

Our Data Table So Far ml NaOH added ph 0.00 ml 1.00 5.00 ml 1.18 10.00 ml 15.00 ml 20.0 ml 25.0 ml 7.00 30.0 ml 35.0 ml 40.0 ml 50.0 ml

Use the same technique to find the ph for 10.00, 15.00, and 20.00 ml of added NaOH ml NaOH added ph 0.00 ml 1.00 5.00 ml 1.18 10.00 ml 15.00 ml 20.0 ml 25.0 ml 7.00 30.0 ml 35.0 ml 40.0 ml 50.0 ml

Use the same technique to find the ph for 10.00, 15.00, and 20.00 ml of added NaOH ml NaOH added ph 0.00 ml 1.00 5.00 ml 1.18 10.00 ml 1.37 15.00 ml 1.60 20.0 ml 1.95 25.0 ml 7.00 30.0 ml 35.0 ml 40.0 ml 50.0 ml

Strong Acid w/ Strong Base Let s look at how we can plot the titration curve for the titration of 25.0 ml of 0.100 M HCl with 0.100 M NaOH. After equivalence is when we add more than 25 ml NaOH calculate [OH - ] by subtracting initial moles of H 3 O + from the number of moles of added OH -, then divide by the total volume Let s look at the ph after I add 30.0 ml of NaOH

Strong Acid w/ Strong Base Let s look at how we can plot the titration curve for the titration of 25.0 ml of 0.100 M HCl with 0.100 M NaOH. Let s look at the ph after adding 30.00 ml of NaOH 0.0300 L NaOH (0.100 mol/l) = 0.00300 mol NaOH Addition 0.00300 OH - mol H 3 -- O + After Initial addition (moles) 0.000500 0.00 mol (0.00250 0.00250 0.00 mol mol Addition mol 0.00300 reacts with mol the H 3 O + -- After addition OH - H 3 O + Initial (moles) 0.00 mol 0.00250 mol present, and then 0.000500 is left)

Strong Acid w/ Strong Base Let s look at how we can plot the titration curve for the titration of 25.0 ml of 0.100 M HCl with 0.100 M NaOH. Let s look at the ph after I add 30.0 ml of NaOH So [OH - ] = 0.000500 mol/ (0.0250L + 0.0300 L) [OH - ] = 0.00909 poh = 2.04 ph = 11.96 OH - H 3 O + Initial (moles) 0.00 mol 0.00250 mol Addition 0.00300 mol -- After addition 0.000500 0.00 mol

Our Data Table So Far ml NaOH added ph 0.00 ml 1.00 5.00 ml 1.18 10.00 ml 1.37 15.00 ml 1.60 20.0 ml 1.95 25.0 ml 7.00 30.0 ml 11.96 35.0 ml 40.0 ml 50.0 ml

Finish the Table J ml NaOH added ph 0.00 ml 1.00 5.00 ml 1.18 10.00 ml 1.37 15.00 ml 1.60 20.0 ml 1.95 25.0 ml 7.00 30.0 ml 11.96 35.0 ml 40.0 ml 50.0 ml

Finish the Table J ml NaOH added ph 0.00 ml 1.00 5.00 ml 1.18 10.00 ml 1.37 15.00 ml 1.60 20.0 ml 1.95 25.0 ml 7.00 30.0 ml 11.96 35.0 ml 12.22 40.0 ml 12.36 50.0 ml 12.52

Adding NaOH to HCl added 25.0 ml added 30.0 35.0 5.0 10.0 25.0 0.100 ml ml mlnaoh NaOH M NaOH HCl 0.00050 0.00100 0.00250mol 0.00200 0.00150 equivalence molnaoh point HCl ph ph==12.22 11.96 1.00 1.18 1.37 7.00 added 40.0 15.0 ml NaOH 0.00100 mol NaOH 0.00150 HCl ph = 12.36 1.60 added 50.0 20.0 ml NaOH 0.00050 mol NaOH 0.00250 HCl ph = 12.52 1.95

Titration Curve:

Strong Acid w/ Strong Base Notes: The ph changes VERY quickly near the equivalence point This means that small amounts of added base cause large changes in ph

Titration of a Strong Base with a Strong Acid

Titration of a Strong Base with a Strong Acid Consider the ph curve for the titration of 100.0 ml of 0.50 M NaOH with 1.0 M HCl OH is in excess before equivalence point H 3 O + is in excess after the equivalence point

Try This: A 50.0-mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M nitric acid. Calculate the ph a. After adding 30.00 ml of HNO 3 b. At the equivalence point Answers: a. poh à 1.30, so ph = 12.70 b. ph = 7.00