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And Acid/Base dilution

Mr. Shields Regents Chemistry U15 L05

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NeutralizationNeutralization

Remember we said previously that one of the Property’s of acids and bases is that they react together

Do you remember what this reaction is called?It’s neutralization. So what exactly is neutralization?

Neutralization occurs when acids and bases react. Thisresults in 2 products – a salt and water. For example:

HNO3 + LiOH LiNO3 + H20

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TitrationTitration

Titrations are carefully “Controlled” Neutralizations.

A titration is used to determine the amount of acid orBase present in an unknown.

- for example: you might want to know how acidic is in a sample of collected lake water.

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Acid-Base TitrationAcid-Base Titration Requires a Requires a standard solutionstandard solution an an acid-acid-

base indicator base indicator and aand a graduated burette graduated burette

Typically the indicator used is Typically the indicator used is PhenolphthaleinPhenolphthalein when titrating when titrating strongstrong acids acids and basesand bases

A Standard solution is an acid or base of A Standard solution is an acid or base of known molar concentrationknown molar concentration..

If titrating an acid we need a standardized base.If titrating an acid we need a standardized base. If titrating a base we need a standardized acidIf titrating a base we need a standardized acid

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Recall Molarity (M) = # moles/liter of solution

For example: If we add 0.5 moles of KOH to 750 ml of water what is the Molarity?

M = 0.5 mol / 0.75L = 0.67M

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TitrationTitration The Standard solution is slowly added The Standard solution is slowly added

to the unknown solution drop by drop.to the unknown solution drop by drop.

As the solutions mix, a neutralization As the solutions mix, a neutralization reaction occurs.reaction occurs.

Eventually, enough standard solution Eventually, enough standard solution is added to is added to neutralizeneutralize the unknown the unknown solution. This is the solution. This is the Equivalence Equivalence pointpoint..

Example: 1 mol H2SO4 + 2 mol KOH 1 mol K2SO4 + 2 mol H20

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1 ml

2

3

4

5

6

1 ml

2

3

4

5

6

Reading ABuret

Initial reading2.4 ml

Final Reading4.8 ml

Volume used = 2.4 ml

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Equivalence pointEquivalence point

Total number of moles of HTotal number of moles of H++ ions ions donated by acid = total number of donated by acid = total number of moles of OHmoles of OH-- donated by the base. donated by the base.

Total moles HTotal moles H++ = total moles OH = total moles OH--

Ex: 50ml 0.5M HCL + 50ml 0.5M KOH have equalAmounts of H+ and OH-

How many moles of HCl and KOH are there? 0.025 mol ea.

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TitrationTitration End-pointEnd-point = point at which indicator changes = point at which indicator changes

color.color.

In titrating an unk. acid phenolphthalein changesIn titrating an unk. acid phenolphthalein changes

from colorless to pinkfrom colorless to pink

(what would the color change be if we were (what would the color change be if we were titrating a base with an acid?)titrating a base with an acid?)

If the indicator is chosen correctly, the end-If the indicator is chosen correctly, the end-point is very close to the equivalence point.point is very close to the equivalence point.

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Titration of a strong acid Titration of a strong acid with a strong basewith a strong base

Volume of 0.100 M NaOH added (ml)

pH

0-

14-

7- Equivalence Pt

Phenolphthalein Phenolphthalein Color change: Color change: 8.28.2 Between pH of 4 and 10, only

a few dropsdrops of base are added.

0 ml

40ml

20 ml

End point

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Calculation of ConcentrationCalculation of ConcentrationThink back to our discussion of solutions. Do you Think back to our discussion of solutions. Do you Recall the equation for determining Molarity byRecall the equation for determining Molarity byDilution?Dilution?

Calculating Molarity after dilution is accomplished veryCalculating Molarity after dilution is accomplished verysimply by using the equation simply by using the equation MM11VV11 = M = M22VV22

For example if M1 = 2M and V1=100ml and it is thenDiluted with 100ml of H20 what is the new Molarity?

2M x 100ml = M2 x 200ml M2= 200/200 = 1M

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Neutralization EquationNeutralization Equation

In neutralization we use a very similar equationIn neutralization we use a very similar equation

##H+H+ (M (MH+H+VVH+ H+ )= #)= #OH- OH- (M(MOH-OH-VVOH-OH-))

MMH+H+ = molarity of H = molarity of H+ #+ # = Subscript in the = Subscript in the MMOH-OH- = molarity of OH = molarity of OH- - acid or base foracid or base for VVH+H+ = volume of H = volume of H+ + H H++ or OH or OH--

VVOH-OH- = volume of OH = volume of OH--

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and Mand MaaVVaa = M = MbbVVbb

True for strong True for strong monomonoprotic acids protic acids andand monomonohydroxy bases Since both hydroxy bases Since both HH++ and OH and OH-- (S#) =1. (S#) =1.

Or strong diprotic Or strong diprotic andand dihydroxy acids dihydroxy acids & bases since both H& bases since both H++ and OH and OH-- (S#) = (S#) = 22

Like HCl & KOH Or HLike HCl & KOH Or H22SOSO4 4 & Ca(OH)Ca(OH)22

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Titration Problem #1Titration Problem #1

In a titration of In a titration of 40.0 mL40.0 mL of a nitric acid of a nitric acid solution, the end point is reached when solution, the end point is reached when 35.0 mL35.0 mL of of 0.100 M NaOH0.100 M NaOH is added. is added.

Calculate the concentration of the nitric Calculate the concentration of the nitric acid solution.acid solution.

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Neutralization ReactionNeutralization Reaction

HNOHNO33 + NaOH + NaOH H H22O + NaNOO + NaNO33

HNOHNO33 is a strong is a strong monoproticmonoprotic acid.acid.

NaOH is a strong NaOH is a strong monohydroxy basemonohydroxy base

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VariablesVariables

MMaa = ? = ? VVaa = 40.0 mL = 40.0 mL MMbb = 0.100 M = 0.100 M VVbb = 35.0 mL = 35.0 mL # = 1 and 1# = 1 and 1

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Plug and ChugPlug and Chug

MMAA (40.0 mL) = (0.100 M ) (40.0 mL) = (0.100 M )(35.0 mL)(35.0 mL)

X = Molarity (M) = .0875 MX = Molarity (M) = .0875 M

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Titration Problem #2Titration Problem #2

What is the concentration of a What is the concentration of a sulfuric acid solutionsulfuric acid solution if if 50.0 mL50.0 mL of a of a 0.250 M KOH0.250 M KOH solution are needed to solution are needed to neutralize neutralize 20.0 mL of the H20.0 mL of the H22SOSO44 solutionsolution of unknown concentration? of unknown concentration?

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Neutralization ReactionNeutralization Reaction

KOH + HKOH + H22SOSO44 H H22O + KO + K22SOSO44

HH22SOSO44 is a strong is a strong DiproticDiprotic acid. acid. KOH is a strong monohydroxy KOH is a strong monohydroxy

base base ##H+H+ = 2, # = 2, #OH-OH- = 1) = 1)

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VariablesVariables

MMaa = ? = ? VVaa = 20.0 mL = 20.0 mL MMbb = 0.250 M = 0.250 M VVbb = 50.0 mL = 50.0 mL SSAA =2 & S =2 & SBB =1 =1

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Plug and ChugPlug and Chug

##AAxMxMAAxVxVAA = # = #BBxMxMBBxVxVBB

2xM2xMAA (20.0 mL) = 1x(0.250 M) (50.0 (20.0 mL) = 1x(0.250 M) (50.0 mL)mL)

MMAA= 0.312 M= 0.312 M