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TRANSCRIPT
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5.1.12
• What is the pH of a 1.0 L 1M HCl solu7on that has reacted with 1.0L of a 1M NaOH solu7on?
• Today – fundamentals of 7tra7on • Calculate expected pH resul7ng from 7tra7on of a strong acid with a strong base
• Explain func7on of indicators
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Indicators
• An indicator is a substance that whose op7cal proper7es are a func7on of pH
• What does that mean? • It changes color in response to changes in pH • Typically an indicator only works in a specific range of pH values
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Indicators and pH
• In an acidic solu7on, a good indicator will act as a Brønsted base, and accept a proton
• This changes its op7cal proper7es, and gives an inves7gator a visual indica*on of the [H+]
• In basic solu7ons, a good indicator will act as a Brønsted acid, and donate H+ ions to OH-‐ present in the solu7on
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indicators
• The pH range an indicator is effec7ve in is limited, called its transi'on interval
• A good indicator at a low pH is one that ionizes well, and so will only accept protons from strong acids
• A good indicator at a higher pH (but s7ll an acidic pH) is one that weakly ionizes, but s7ll strongly enough so that it can accept protons from weak acids
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Examples of indicators
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Role of indicators
• Not to tell what the pH “is” of a solu7on (for that we use a pH meter, which measures voltage between electrodes placed in the solu7on)
• Indicator is tell us the equivalence point when an acid and base react
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Equivalence point
• The is the point at which the concentra7on of [H+] equals the concentra7on of protonated conjugate
• We use this representa7on HA --- H+ + A-
• Where HA is the Brønsted acid, and A-‐ is the conjugate base
• The equivalance point is when [A-‐] = [HA] • What are these values for water?
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7tra7on
• The process of determining an acids equivalence point by slowly adding a base in the presence of an indicator
• As base is added, pH changes – why?
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Acid equilibrium constant
• Really, this is only useful for weak acids (that do not ionize completely)
• Symbolized by Ka • This is the pH at which [H3O+][A-‐] [HA]
These values are molar concentra7on numbers, just like pH we can write this figure as an inverse log
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5.1.12
• What is [H+] when the pH is 6.32?
• Today – titrations
• Determine the molarity of a strong acid solution titrated to equivalence point by a strong base
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What 7tra7on looks like
An indicator is used to show you when the equivalence
point is reached
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diagrams
• Visual representa7on of a weakly acidic solu7on
• Strongly acidic solu7on • Titra7on with a strong base • Ques7ons: what happens to concentra7on of [HA] for a weak acid as it is 7trated with a strong base? [H+]? [A-‐]?
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Titra7on calcula7ons
• We begin with this formula: • Mbase X Vbase = Macid X V acid • What does this look like? • We also need to consider mole ra7os, which means we need to begin with a balanced chemical reac7on for the 7tra7on
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Example
• In a 7tra7on, 27.4 ml of a 0.0154 M Ba(OH)2 is added to a 20.0 ml sample of HCl of unknown concentra7on un7l an equivalence point is reached. What was the molarity of the HCl? – Step 1: write out balanced equa7on – Step 2: MbaseVbase = MacidVacid – Step 3: determine mole ra7o from reac7on – Step 4: mul7ply known M by mole ra7o so units cancel, remaining units will be in terms of molarity of unknown
• Solu7on on next slide…
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solu7on
• Ba(OH)2 + 2HCl BaCl2 + 2H2O • Need common units for volume (because M is moles/liter), so convert volume units to L
• Mole ra7o 2 mol HCl/1 mol Ba(OH)2 0.0154 mol Ba(OH)2/L X 1L/1000ml = 4.22X10-‐4 mol Ba(OH)2 2 mol HCl/1 mol Ba(OH)2 X 4.22X10-‐4 mol Ba(OH)2
= 8.44 X 10-‐4 mol HCl
8.44 X 10-‐4 mol HCl X 1000ml/L = 4.22 X 10-‐2 mol HCl
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What does the answer mean?
• In that case, you “didn’t know” the ini7al molarity of the HCl solu7on, so the calcula7on allowed you to find that
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Addi7onal examples
• A 15.5 ml sample of 0.215 M KOH solu7on required 21.2 ml of aqueous ace7c acid solu7on to reach equivalence. What was the molarity of the ace7c acid solu7on?
• By 7tra7on, a 17.6 ml aqueous H2SO4 solu7on neutralized 27.4 ml of 0.0165 M LiOH. What was the molarity of the acid?
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What does this show?
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Ka
• Defined is [H3O+][A-‐] [HA]
• pKa = -‐logKa
• Describes the pH value when
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Review: conjugate bases
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5.4.12
• What happens to rela7ve ra7o of acid to conjugate base in a 7tra7on?
• HW – page 523 # 15, 16, 21, 36
• Predict equivalence point for 7tra7ons • Explain Ka and how it used
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Titra7on curve
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Other 7tra7ons
• We’ve looked at strong acid with a strong base – equivalence point will be at about pH 7.0
• Strong acid + weak base – equivalence point will be < 7; why?
• Weak acid + strong base – equivalence point will be > 7; why?
• What about weak acid + weak base? depends on Ka of each
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Ka
• Defined is [H3O+][A-‐] [HA]
• What would a LARGE Ka mean? • What would a small Ka mean?
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Ka problems
• What would the pH be of a 0.12 M ace7c acid solu7on (Ka of ace7c acid is 1.74 X 10-‐5)
• What does that Ka tell you? • Set up: ka = ([H3O+][A-‐])/[HA] • [H3O+] and [A-‐] will always be equal! Why? • Simplify [H3O+][A-‐] to [H+]2 • Rearrange to get [H+]2 = Ka[HA] • So [H+] is the square root of Ka[HA], now plug in what you
know: Ka and [HA] • [H+] = sqrt(1.74X10-‐5 X 0.12 M) • = 1.44 X 10-‐3 • So pH = 2.8
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Kb • Example reac7on: NH3 + H2O NH4+ + OH-‐
• So we have Kb = [HB+][OH-‐]/[B]
• The [dissocia7on products] divided by [non-‐dissociated base]
• This describes how many hydroxide ions are produced from the dissocia7on of a base in water
• Higher Kb = more OH-‐ • This is a way to find pOH! • So rela7vely what would the pH be of a solu7on with of substance
with a high Kb?
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Example of a Kb problem
• What is pH of a 0.100 M solu7on of ammonia (Kb = 1.77 X 10-‐5)
• Strategy: same as with Kb, we are considering a base that dissociates and gives you a single OH-‐, so [BH+][OH-‐] is essen7ally [OH-‐]2
• Rearranging Kb equa7on looks just like Ka: • You will find [OH-‐] = sqrtKb X B • Plug in and solve: • [OH-‐] = sqrt {1.77X10-‐5 X 0.100 M} • Remember this lets you find pOH, so subtract this from 14 and you have pH
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5.7.12
• What does Ka tell you about an acid?
• Lab tomorrow – equip not available today
• Today: prac7ce 7tra7on and equilibrium calcula7ons instead
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Review: Ka • Ka = [H3O+][A-‐] [HA] BUT -‐-‐-‐ this is making an assump7on – the only source of H+
ions is the acid dissolved in water. Is there another source of H+?
YES – the water! Also – what happens to the molarity of HA when it is
dissolved? Does it stay the same? NO! it decreases by [A-‐] (this is a small number for weak
acids!) So is it possible to accurately know pH to many decimal
places?
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Example – two solu7ons
• A 0.120 M solu7on of a generic weak acid (HA) has a pH of 3.26. Determine the Ka.
• Solu7on 1: use the equa7on you know, solve for Ka
• Solu7on 2: use this equa7on (which in some books/websites is the proper equa7on)
• Ka = [H+][A-‐] [HA]-‐[A-‐] What’s the difference? Does the method maxer? Some%mes!
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Today – green books!!
• Page 285 • Look at 1, 2, 3, 5 and 6
• P 293 1&2 • P296 #1 • P299 #1