Titration

• Indicators & pH Meters: The pH range over which an indicator changes color is its transition intervaltransition interval. pH meters measure the pH using electrodes.

• TitrationTitration is the process of adding an acid to base until it is neutralized. (or vice versa)

• Indicators are used to show when neutralizationneutralization is reached.

More On Titrations

• Titrations are done to find the concentrationconcentration of the unknown acid or base.

• The correct indicator needs to be chosen so that it changes color right where acid = baseacid = base or very close to this point.

• The place where the indicator changes color is the end point.end point.

• The place where acid = base is the equivalence equivalence point.point. – Titrations may be done on compounds that are not acids

& bases if they react so that the two factors are in chemically equivalent amounts at the equivalence point & there is an indicator with an appropriate endpoint.

For a titration with a strong acid and a strong base the equivalence point occurs at approximately a pH of 7pH of 7.

For a titration with a weak acid and a strong base the equivalence point occurs above aabove a pH of 7pH of 7.

Performing Titrations

• STEPS:• 1.) Use burets to accurately measure acid & base

amounts. • 2.) Add one to the other until the indicator just

barely changes colorcolor. • If the acid & base involved are both monoprotic,monoprotic,

M1V1=M2V2 ( or MHVH=MOHVOH ) can be used to find the unknown molarity.

• If the acid & base involved have different amounts of H+ and OH-, then the balanced equationbalanced equation will have to be used along with dimensional analysis. (p502)

Sample Problem

• If 20.0 mL of 0.0100 M aqueous HCl is required to neutralize 30.0 mL of an aqueous solution of NaOH, determine the molarity of the NaOH solution.

• MMHHVVHH=M=MOHOHVVOHOH

• (0.0100M) (20.0 mL) = (M(0.0100M) (20.0 mL) = (MOHOH)(30.0mL))(30.0mL)

• MMOH OH = 6.67 x 10= 6.67 x 10-3-3 M NaOH M NaOH

Try it Out!

• In a titration, 27.4 mL of 0.154 M LiOH is added to a 25.0 mL HCl solution of unknown concentration. What is the molarity of the solution?

• MMHHVVHH=M=MOHOHVVOHOH

• (M(MHH)(25.0mL) = (27.4 mL)(0.154M) )(25.0mL) = (27.4 mL)(0.154M)

• MMHH= 0.169 M HCl= 0.169 M HCl

Your Turn!• Suppose 20.0 mL of 0.100 M aqueous Ca(OH)2 is required to

neutralize 12.0 mL of an aqueous solution of HCl. What is the molarity of the HCl solution.

• Ca(OH)Ca(OH)22 + 2HCl + 2HCl 2H 2H22O + CaClO + CaCl22

• Moles H+ = Moles OH- at equiv pointMoles H+ = Moles OH- at equiv pointCalculate moles of OH- (Molarity x Volume)Calculate moles of OH- (Molarity x Volume)

0.100 mols/L x 0.020 L = 0.0020 moles Ca(OH)0.100 mols/L x 0.020 L = 0.0020 moles Ca(OH) 2 2

0.00200 mol Ca(OH)0.00200 mol Ca(OH)2 2 x x 2 mol HCl 2 mol HCl = 0.0040 mol HCl= 0.0040 mol HCl

1 mol Ca(OH)1 mol Ca(OH)2 2

Molarity =Molarity = 0.004 mols HCl 0.004 mols HCl = 0.33 M HCl = 0.33 M HCl0.012 L0.012 L

Another One!• By titration, 17.6 mL of aqueous H2SO4 neutralized 27.4 mL

of an aqueous solution of 0.0165 M KOH. What is the molarity of the acid solution?

• HH22SOSO44 + 2KOH + 2KOH 2H 2H22O + KO + K22SOSO44

Moles H+ = Moles OH- at equiv pointMoles H+ = Moles OH- at equiv pointCalculate moles of OH- (Molarity x Volume)Calculate moles of OH- (Molarity x Volume)0.0165 moles/L x 0.0274 L = 0.000452 moles KOH0.0165 moles/L x 0.0274 L = 0.000452 moles KOH0.000452 moles KOH x 0.000452 moles KOH x 1 mol H1 mol H22SOSO44 = = 0.000226 mol H 0.000226 mol H22SOSO44

2 mol KOH2 mol KOH

M =M = 0.000226 mols 0.000226 mols HH22SOSO44 = 0.0128 M = 0.0128 M HH22SOSO44

0.0176 L0.0176 L

Last One!• By titration, 33.3 mL of aqueous H3PO4 neutralized 55.5 mL of an aqueous

solution of 0.444 M Ba(OH)2. What is the molarity of the acid solution?

2H2H33POPO44 + 3Ba(OH) + 3Ba(OH)22 6H 6H22O + BaO + Ba33(PO(PO4 4 ))22

Moles H+ = Moles OH- at equiv pointMoles H+ = Moles OH- at equiv point

Calculate moles of Ba(OH)Calculate moles of Ba(OH)22 (Molarity x Volume) (Molarity x Volume)

0.444 moles/L x 0.0555 L = 0.0246 moles 0.444 moles/L x 0.0555 L = 0.0246 moles

0.0246 moles Ba(OH)0.0246 moles Ba(OH)2 2 x x 2 mol H2 mol H33POPO44 = = 0.0164 mol H 0.0164 mol H33POPO44

3 mol Ba(OH)3 mol Ba(OH)22

M =M = 0.0164 mols 0.0164 mols HH33POPO44 = 0.493 M = 0.493 M HH33POPO44

0.0333 L0.0333 L

What was the longest recorded time that a single person had the hiccups?

68 Years!

Charles Osborne had the hiccups from the age of

28 to 96. He hiccupped an estimated 430 million times!