ACID-BASE TITRATION AND PH
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1 ACID-BASE TITRATION AND PH
2 Section 1 Aqueous Solutions and the Concept of ph
3 Hydronium and Hydroxide Ions Acids and bases form hydroxide and hydronium ions These ions are not the only ones in an aqueous solution Hydronium and hydroxide are also produced by water
4 Self-Ionization of Water Water can undergo self-ionization Self-ionization of water à two water molecules make a hydronium and hydroxide ion by transfer of a proton
5 Conductivity measurements show that concentrations of H 3 O + and OH - in pure water are each only 1.07 x 10-7 mol/l of water at 25 C That means in 10,000 L of water, less than 1 drop would be ionized
6 Standard notation represents concentration in moles per liter Formula for ion or molecule in brackets [ ] Ex. [H 3 O + ] is concentration of hydronium in moles per liter In water at 25 C [H 3 O + ] = 1.0 x 10-7 M and [OH - ] = 1.0 x 10-7 M
7 Mathematical product of [H 3 O + ] and [OH - ] remains constant in water and dilute aqueous solutions at constant temperature This constant is called the ionization constant of water, K w K w = [H 3 O + ][OH - ] K w = (1.0 x 10-7 M)(1.0 x 10-7 M) = 1.0 x M 2
8 Ionization of water increases as temperature increases So, K w increases as temp increases At any given temp, K w is always a constant value
9 Neutral, Acidic, and Basic Solutions In pure water, [H 3 O + ] = [OH - ] It is neutral Any solution where [H 3 O + ] = [OH - ] is neutral Remember acids increase [H 3 O + ] in aqueous solutions Acidic solutions: [H 3 O + ] > [OH - ] Basic solutions: [H 3 O + ] < [OH - ]
10 Calculating [H 3 O + ] and [OH - ] Strong acids and bases are considered completely ionized or dissociated in weak aqueous solutions 1 mol NaOH yields 1 mol OH - in aqueous solution So 1.0 x 10-2 M NaOH has [OH - ] of 1.0 x 10-2 M
11 Is greater than 1.0 x 10-7 M Solution is basic b/c K w is relatively constant at 1.0 x M 2 at ordinary room temps, concentration of either ion can be calculated if concentration of 1 of them is known
12 Now consider a 1.0 x 10-4 M H 2 SO 4 solution Remember H 2 SO 4 is diprotic Acidic or basic?
13 Calculate [OH-] -11
14 The hydroxide ion concentration of an aqueous solution is 6.4 x 10-5 M. What is the hydronium concentration? (No, it s not in your sheet) 1.6 x M
15 Calculate the H 3 O + and OH - concentrations in a 7.50 x 10-4 M solution of HNO 3, a strong acid. (once again, NOT in your sheet ;-) [H 3 O + ] = 7.50 x 10-4 M [OH - ] = 1.33 x M
16 Calculate [H 3 O + ] and [OH - ] of the following solutions. They are all strong acids/bases a M sodium hydroxide b M sulfuric acid c M lithium hydroxide d M nitric acid e M calcium hydroxide f M perchloric acid
17 a M sodium hydroxide [OH - ] = 0.05 M [H 3 O + ] = 2 x M b M sulfuric acid [OH - ] = 2.0 x M [H 3 O + ] = 5.0 x 10-3 M
18 c M lithium hydroxide [OH - ] = M [H 3 O + ] = 7.7 x M d M nitric acid [OH - ] = 6.67 x M [H 3 O + ] = M
19 e M calcium hydroxide [OH - ] = M [H 3 O + ] = 2.5 x M f M perchloric acid [OH - ] = 2.56 x M [H 3 O + ] = M
20 The ph Scale ph stands for French words pourvoir hydrogène meaning hydrogen power ph à the negative of the common logarithm of the hydronium ion concentration, [H 3 O + ] ph = -log [H 3 O + ]
21 Common logarithm of a number is the power to which 10 must be raised to equal the number Neutral solution at room temp has [H 3 O + ] of 1 x 10-7 M Logarithm of 1 x 10-7 is -7.0
22 poh à the negative of the common logarithm of the hydroxide ion concentration [OH - ] poh = -log [OH - ] Remember K w = [H 3 O + ][OH - ] = 1.0 x Negative log = 14.0 So ph + poh = 14.0 at room temp
23 ph of acidic solution at 25 C with [H 3 O + ] of 1 x 10-6 M is 6.0 ph is less than 7
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25 Calculations Involving ph If either [H 3 O + ] or ph is known the other can be calculated Significant figures involving ph must be handled carefully b/c ph represents a log, the number to the left of the decimal only locates the decimal point Not included when counting significant figures
26 There must be as many sig figs to the right of the decimal as there are in the number whose log was found Ex. [H 3 O + ] = 1 x 10-7 One sig fig So, ph or log must have one digit to the right of the decimal point So, ph = 7.0
27 Calculating ph from [H 3 O + ] What is the ph of a 1.0 x 10-3 M NaOH solution?
28 1. Analyze Given: Identity and concentration of solution (1.0 x 10-3 M NaOH) Unknown: ph of solution
29 2. Plan [base] à [OH - ] à [H 3 O + ] à ph
30 3. Compute
31 What is the ph of a solution with the following hydroxide ion concentrations? a. 1 x 10-5 M 9.0 b. 2 x 10-8 M 6.7 c x M 3.46
32 Using a Calculator to Calculate ph from [H 3 O + ] Some problems involve [H 3 O + ] not equal to integral powers of 10 These require a calculator Ex. What is the ph of a solution if the [H 3 O + ] is 3.4 x 10-5 M?
33 What is the ph of a solution if the [H 3 O + ] is 3.4 x 10-5 M? The only difference is that you will determine the log of 3.4 x 10-5 M using a calculator ph = -log [H 3 O + ] = -log (3.4 x 10-5 ) = 4.47
34 1. (-) 2. Log 3. ( Shift 6. 10x 7. (-) ) 10. =
35 What are the poh values of solutions with the following [H 3 O + ]? a x M 1.40 b. 4.3 x 10-3 M 11.6 c. 9.1 x 10-6 M 8.96 d M 12.8
36 A solution is prepared by dissolving 3.50 g of sodium hydroxide in water and adding water until the total volume of the solution is 2.50 L. What are [OH - ] and [H 3 O + ]? [OH - ] = M [H 3 O + ] = 2.86 x M
37 Calculating [H 3 O + ] and [OH - ] from ph You already know ph = -log [H 3 O + ] Base of common logs is 10 So antilog of common log is 10 raised to that number Log[H 3 O + ] = antilog (-ph) [H 3 O +] = 10 -ph
38 Determine the [H 3 O + ] of an aqueous solution with ph of 4.0 ph = -log [H 3 O + ] Log [H 3 O + ] = -ph [H 3 O + ] = antilog (-ph) [H 3 O + ] = 1 x 10 -ph [H 3 O + ] = 1 x 10-4 M
39 What are the poh and [OH - ] of a solution with a ph of 8.92? poh = 5.08 [OH - ] = 8.3 x 10-6 M
40 A solution of NH3 has a ph of Calculate [H 3 O + ] and [OH - ]. [H 3 O + ] = 1.0 x M [OH - ] = 1.0 x 10-3 M
41 Determining ph and Titrations Section 2
42 Indicators and ph meters Acid-base indicators à compounds whose colors are sensitive to ph They change color b/c they re either weak acids or weak bases In solution, equilibrium of indicator that is weak acid can be represented by In- is the symbol of the anion part of the indicator
43 Colors that indicator displays result from the fact that HIn and In - are different colors
44 In acidic solutions, any In- ions that are present act as Brønsted-Lowry bases and accept protons from acid Indicator is then present in mostly nonionized form, Hin Indicator will have acidindicating color (red)
45 In basic solutions, OHions from base combine with H+ made by indicator Indicator molecules ionize to offset loss of H+ ions Indicator present mostly in anion form, In- Solution now shows base-indicating color, blue
46 Indicators come in different colors Exact ph range also varies Transition interval à ph range over which an indicator changes color
47 Indicators that change color at ph lower than 7 (methyl orange) are stronger acids than other types of indicators They ionize more than other indicators
48 The In- anions made by these are weaker Bronsted bases and are less likely to accept protons from any acid being tested So they don t shift to their nonionized (HIn) form unless concentration of H+ fairly high Color transition happens at lower ph
49 In contrast, indicators that go through transition at higher ph (phenolphthalein) are weaker acids
50 Universal indicators made by mixing several different indicators Paper soaked in universal indicator à ph paper
51 ph meter à determines the ph of a solution by measuring voltage between two electrodes that are placed in solution Voltage changes as hydronium ion concentration in solution changes
52 Titration Neutralization reactions happen between acids and bases OH - gets a proton from H 3 O + forming 2 H 2 O
53 This shows that 1 mol hydronium (19.0g) and 1 mol hydroxide (17.0g) are chemically equal masses They combine in 1:1 ratio Neutralization occurs when hydronium and hydroxide ions are supplied in equal numbers by reactants
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55 1 L of 0.10 M HCl contains 0.1 mol H 3 O + Suppose 0.10 mol solid NaOH is added to 1 L of 0.10 M HCl NaOH dissolves and supplies 0.10 mol OH - HCl and NaOH are in chemically equal amounts H 3 O + and OH - combine until product [H 3 O + ] [OH - ] returns to 1.0 x M 2
56 Because acids and bases react, progressive addition of acid to base (or base to acid) can be used to compare concentrations of acid and base Titration à the controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration
57 Equivalence Point Equivalence point à the point at which two solution used in a titration are present in chemically equal amounts Indicators and ph meters used to determine the equivalence point ph meter shows large voltage change at EP End point à the point in a titration at which an indicator changes color
58 Some indicators, like litmus, change color at about ph 7 BUT the color change interval for litmus is wide, ph Makes it difficult to determine accurate ph Bromothymol blue is better ph Good for strong-acid/strong-base titrations
59 Indicators that change at ph lower than 7 useful to determine EP of strong-acid/weak-base titrations Methyl orange, ex. EP of strong-acid/weak-base titration is acidic b/c salt formed is a weak acid So, salt solution has ph lower than 7
60 Indicators changing color at ph higher than 7 good for weak-acid/strong-base titrations Phenolphthalein, ex. These reactions make salt solutions ph higher than 7 This is b/c salt formed is weak base
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62 As base is added, ph changes from low value to high one Change in ph slow at first, then quickly through EP Then slows down as solution becomes more acidic
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64 Molarity and Titration If the concentration of 1 solution is known, the concentration of the other solution in a titration can be calculated Standard solution à solution that contains the precisely known concentration of a solute Often called known solution
65 To be certain of concentration of known solution, that solution must be compared with solution of a primary standard Primary standard à a highly purified solid compound used to check the concentration of the known solution in a titration
66 Known solution prepared first Volume adjusted to give roughly the desired concentration Concentration then determined more precisely by titrating it with a measured quantity of primary standard
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73 Suppose 20.0 ml of 5.0 x 10-3 M NaOH required to reach end point in titration of 10.0 ml of HCl of unknown concentration How can this data be used to determine molarity of acidic solution?
74 Start with balanced neutralization reaction equation From equation, determine chemically equivalent amounts of HCl and NaOH
75 Calculate number of moles of NaOH used in titration b/c 1 mol NaOH needed to neutralize 1 mol HCl, amount of HCl in titration must be 1.0 x 10-4 mol
76 This amount of acid must be in 10.0 ml of HCl used for titration Now calculate molarity
77 Procedure 1. Start with balanced equation for neutralization reaction, and determine chemically equivalent amounts of acid and base. 2. Determine number of moles of acid/base from known solution used in titration. 3. Determine moles of solute of unknown. 4. Determine molarity of unknown solution.
78 Sample Problem In a titration, 27.4 ml of M Ba(OH) 2 is added to a 20.0 ml sample of HCl solution of unknown concentration. What is the molarity of the acid solution?
79 1. Analyze Given: Volume and concentration of known solution 27.4 ml of M Ba(OH) 2 Volume of unknown HCl solution 20.0 ml Unknown: Molarity of acid solution
80 2. Plan 1. Balanced neutralization equation à chemically equivalent amounts Ba(OH) 2 + 2HCl à BaCl 2 + 2H 2 O 1 mol 2 mol 1 mol 2 mol Mole ratio is 1 mol Ba(OH) 2 for every 2 mol HCl
81 2. Volume known basic solution used (ml) à amount of base used (mol)
82 3. Mole ratio, moles of base used à moles of acid used from unknown solution
83 4. Volume of unknown, moles of solute in unknown à molarity of unknown
84 A student titrates a ml sample of a solution of HBr with unknown molarity. The titration requires ml of a M solution of NaOH. What is the molarity of the HBr solution? M HBr
85 Vinegar can be assayed to determine its acetic acid content. Determine the molarity of acetic acid in a ml sample of vinegar that requires ml of a M solution of NaOH to reach the equivalence point M
86 A ml sample of a solution of Sr(OH) 2 is titrated to the equivalence point with ml of M HCl. What is the molarity of the Sr(OH) 2 solution? M
87 A ml sample of ammonia solution is titrated to the equivalence point with ml of a ml sulfuric acid solution. What is the molarityof the ammonia solution? 1.26 M
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