ACID-BASE TITRATION AND PH

Aqueous Solutions and the Concept of pH

Section 1

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

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

¨ Conductivity measurements show that concentrations of H3O+ 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

¨ Standard notation represents concentration in moles per liter

¨ Formula for ion or molecule in brackets [ ] ¨ Ex. [H3O+] is concentration of hydronium in moles

per liter¨ In water at 25°C [H3O+] = 1.0 x 10-7 M and [OH-]

= 1.0 x 10-7 M

¨ Mathematical product of [H3O+] and [OH-] remains constant in water and dilute aqueous solutions at constant temperature

¨ This constant is called the ionization constant of water, Kw

Kw = [H3O+][OH-]

Kw = (1.0 x 10-7 M)(1.0 x 10-7 M) = 1.0 x 10-14 M2

¨ Ionization of water increases as temperature increases

¨ So, Kw increases as temp increases

¨ At any given temp, Kw is always a constant value

Neutral, Acidic, and Basic Solutions

¨ In pure water, [H3O+] = [OH-]¨ It is neutral¨ Any solution where [H3O+] = [OH-] is neutral¨ Remember acids increase [H3O+] in aqueous

solutions¨ Acidic solutions: [H3O+] > [OH-]¨ Basic solutions: [H3O+] < [OH-]

Calculating [H3O+] 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

¨ Is greater than 1.0 x 10-7 M ¨ Solution is basic¨ b/c Kw is relatively constant at 1.0 x 10-14 M2 at

ordinary room temps, concentration of either ion can be calculated if concentration of 1 of them is known

¨ Now consider a 1.0 x 10-4 M H2SO4 solution¨ Remember H2SO4 is diprotic

¨ Acidic or basic?

¨ Calculate [OH-]

-11

¨ The hydroxide ion concentration of an aqueous solution is 6.4 x 10-5 M. What is the hydroniumconcentration? (No, it’s not in your sheet)

¨ 1.6 x 10-10 M

¨ Calculate the H3O+ and OH- concentrations in a 7.50 x 10-4 M solution of HNO3, a strong acid. (once again, NOT in your sheet ;-)

¨ [H3O+] = 7.50 x 10-4 M¨ [OH-] = 1.33 x 10-11 M

¨ Calculate [H3O+] and [OH-] of the following solutions. They are all strong acids/bases

¨ a. 0.05 M sodium hydroxide¨ b. 0.0025 M sulfuric acid¨ c. 0.013 M lithium hydroxide¨ d. 0.150 M nitric acid¨ e. 0.0200 M calcium hydroxide¨ f. 0.390 M perchloric acid

¨ a. 0.05 M sodium hydroxide¨ [OH-] = 0.05 M¨ [H3O+] = 2 x 10-13 M

¨ b. 0.0025 M sulfuric acid¨ [OH-] = 2.0 x 10-12 M¨ [H3O+] = 5.0 x 10-3 M

¨ c. 0.013 M lithium hydroxide¨ [OH-] = 0.013 M¨ [H3O+] = 7.7 x 10-13 M

¨ d. 0.150 M nitric acid¨ [OH-] = 6.67 x 10-14 M¨ [H3O+] = 0.150 M

¨ e. 0.0200 M calcium hydroxide¨ [OH-] = 0.0400 M¨ [H3O+] = 2.5 x 10-13 M

¨ f. 0.390 M perchloric acid¨ [OH-] = 2.56 x 10-14 M¨ [H3O+] = 0.390 M

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, [H3O+]

¨ pH = -log [H3O+]

¨ Common logarithm of a number is the power to which 10 must be raised to equal the number

¨ Neutral solution at room temp has [H3O+] of 1 x 10-

7 M¨ Logarithm of 1 x 10-7 is -7.0

¨ pOH à the negative of the common logarithm of the hydroxide ion concentration [OH-]

¨ pOH = -log [OH-]¨ Remember Kw = [H3O+][OH-] = 1.0 x 10-14

¨ Negative log = 14.0¨ So pH + pOH = 14.0 at room temp

¨ pH of acidic solution at 25°C with [H3O+] of 1 x 10-6 M is 6.0

¨ pH is less than 7

Calculations Involving pH

¨ If either [H3O+] 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

¨ There must be as many sig figs to the right of the decimal as there are in the number whose log was found

¨ Ex. [H3O+] = 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

Calculating pH from [H3O+]

¨ What is the pH of a 1.0 x 10-3 M NaOH solution?

1. Analyze

¨ Given:

¨ Identity and concentration of solution (1.0 x 10-3 M NaOH)

¨ Unknown:¨ pH of solution

2. Plan

¨ [base] à [OH-] à [H3O+] à pH

3. Compute

¨ 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. 2.90 x 10-11 M¨ 3.46

Using a Calculator to Calculate pH from [H3O+]

¨ Some problems involve [H3O+] not equal to integral powers of 10

¨ These require a calculator

¨ Ex. What is the pH of a solution if the [H3O+] is 3.4 x 10-5 M?

What is the pH of a solution if the [H3O+] 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 [H3O+] ¨ = -log (3.4 x 10-5)¨ = 4.47

1. (-)2. Log3. (4. 3.45. Shift6. 10x7. (-)8. 59. )10. =

¨ What are the pOH values of solutions with the following [H3O+]?

¨ a. 2.51 x 10-13 M¨ 1.40¨ b. 4.3 x 10-3 M¨ 11.6¨ c. 9.1 x 10-6 M¨ 8.96¨ d. 0.070 M¨ 12.8

¨ 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 [H3O+]?

¨ [OH-] = 0.0350 M¨ [H3O+] = 2.86 x 10-13 M

Calculating [H3O+] and [OH-] from pH

¨ You already know¨ pH = -log [H3O+]¨ Base of common logs is 10¨ So antilog of common log is 10 raised to that

number¨ Log[H3O+] = antilog (-pH)¨ [H3O+] = 10-pH

¨ Determine the [H3O+] of an aqueous solution with pH of 4.0

¨ pH = -log [H3O+] ¨ Log [H3O+] = -pH¨ [H3O+] = antilog (-pH)¨ [H3O+] = 1 x 10-pH

¨ [H3O+] = 1 x 10-4 M

¨ 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

¨ A solution of NH3 has a pH of 11.00. Calculate [H3O+] and [OH-].

¨ [H3O+] = 1.0 x 10-11 M¨ [OH-] = 1.0 x 10-3 M

Section 2

Determining pH and Titrations

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

¨ Colors that indicator displays result from the fact that HIn and In- are different colors

¨ 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 nonionizedform, Hin

¨ Indicator will have acid-indicating color (red)

¨ In basic solutions, OH-ions 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

¨ Indicators come in different colors¨ Exact pH range also varies¨ Transition interval à pH range over which an

indicator changes color

¨ 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

¨ The In- anions made by these are weaker Bronstedbases 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

¨ In contrast, indicators that go through transition at higher pH (phenolphthalein) are weaker acids

¨ Universal indicators made by mixing several different indicators

¨ Paper soaked in universal indicator à pH paper

¨ pH meter àdetermines 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

Titration

¨ Neutralization reactions happen between acids and bases

¨ OH- gets a proton from H3O+ forming 2 H2O

¨ 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

¨ 1 L of 0.10 M HCl contains 0.1 mol H3O+

¨ 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¨ H3O+ and OH- combine until product [H3O+] [OH-]

returns to 1.0 x 10-14 M2

¨ 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

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

¨ Some indicators, like litmus, change color at about pH 7

¨ BUT the color change interval for litmus is wide, pH 5.5-8.0

¨ Makes it difficult to determine accurate pH¨ Bromothymol blue is better¨ pH 6.0-7.6¨ Good for strong-acid/strong-base titrations

¨ 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

¨ 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

¨ 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

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”

¨ 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

¨ 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

¨ 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?

¨ Start with balanced neutralization reaction equation¨ From equation, determine chemically equivalent

amounts of HCl and NaOH

¨ 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

¨ This amount of acid must be in 10.0 mL of HCl used for titration

¨ Now calculate molarity

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.

Sample Problem

¨ In a titration, 27.4 mL of 0.0154 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?

1. Analyze

¨ Given:

¨ Volume and concentration of known solution¨ 27.4 mL of 0.0154 M Ba(OH)2¨ Volume of unknown HCl solution¨ 20.0 mL

¨ Unknown:

¨ Molarity of acid solution

2. Plan

1. Balanced neutralization equation à chemically equivalent amounts

Ba(OH)2 + 2HCl à BaCl2 + 2H2O1 mol 2 mol 1 mol 2 mol

Mole ratio is 1 mol Ba(OH)2 for every 2 mol HCl

2. Volume known basic solution used (mL) à amount of base used (mol)

¨ 3. Mole ratio, moles of base used à moles of acid used from unknown solution

4. Volume of unknown, moles of solute in unknown àmolarity of unknown

¨ A student titrates a 20.00 mL sample of a solution of HBr with unknown molarity. The titration requires 20.05 mL of a 0.1819 M solution of NaOH. What is the molarity of the HBr solution?

¨ 0.1824 M HBr

¨ Vinegar can be assayed to determine its acetic acid content. Determine the molarity of acetic acid in a 15.00 mL sample of vinegar that requires 22.70 mL of a 0.550 M solution of NaOH to reach the equivalence point.

¨ 0.832 M

¨ A 20.00 mL sample of a solution of Sr(OH)2 is titrated to the equivalence point with 43.03 mL of 0.1159 M HCl. What is the molarity of the Sr(OH)2 solution?

¨ 0.1247 M

¨ A 35.00 mL sample of ammonia solution is titrated to the equivalence point with 54.95 mL of a 0.400 mL sulfuric acid solution. What is the molarity of the ammonia solution?

¨ 1.26 M