edpH

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Acid and Bases

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Acid and Bases

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Acid and Bases

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Acids

Have a sour taste. Vinegar is a solution of acetic acid. CitrusHave a sour taste. Vinegar is a solution of acetic acid. Citrusfruits contain citric acid.fruits contain citric acid.

React with certain metals to produce hydrogen gasReact with certain metals to produce hydrogen gas ..

React with carbonates and bicarbonates to produce carbonReact with carbonates and bicarbonates to produce carbondioxide gasdioxide gas

Have a bitter taste.Have a bitter taste.

Feel slippery. Many soaps contain bases.Feel slippery. Many soaps contain bases.

Bases

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S ome Properties of Acids

Produce H + (as H 3O +) ions in water (the

hydronium ion is a hydrogen ion attached to

a water molecule)

Taste sour

Corrode metals

Electrolytes

React with bases to form a salt and water

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Acid Nomenclature

Flowchart

hydro- prefix-ic ending

2 elements

-ate endingbecomes-ic ending

-ite endingbecomes

-ous ending

no hydro- prefix

3 elements

A CIDSstart with 'H'

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HBr (aq)

H2CO 3

H2SO 3

hydro bromic acid

carbon ic acid

sulfur ous acid

Acid Nomenclature Review

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S ome Properties of Bases

Produce OH - ions in water

Taste bitter, chalky

A re electrolytes

Feel soapy, slippery

React with acids to form salts and water

pH greater than 7

Turns red litmus paper to blue ³ Basic

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S ome Common Bases

N aOH sodium hydroxide

KOH potassium hydroxide liquid soap

Ba(OH) 2 barium hydroxide stabilizer for plastics

Mg(OH) 2 magnesium hydroxide ³MOM´ Milk of

magnesia

A l(OH) 3 aluminum hydroxide Maalox (antacid)

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pH of CommonS ubstances

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T he pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers,

we just use the NEGA T IVEpower of 10 on the Molarityof the H + (or O H-) ion.

Under 7 = acid7 = neutral

O ver 7 = base

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pHpH [H +][H+] [OH -][OH -] pOHpOH

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pH testingT here are several ways to test pH ± Blue litmus paper (red = acid) ± Red litmus paper (blue = basic)

± pH paper (multi-colored) ± pH meter (7 is neutral, <7 acid, >7

base)

± Universal indicator (multi-colored) ± Indicators like phenolphthalein ± Natural indicators like red

cabbage, radishes

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Paper testingPaper tests like litmus paper and pHpaper ± Put a stirring rod into the solution

and stir.

±T

ake the stirring rod out, and placea drop of the solution from the endof the stirring rod onto a piece of thepaper

± Read and record the color change.Note what the color indicates.

± You should only use a small portionof the paper. You can use onepiece of paper for several tests.

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pH meter

T ests the voltage of the electrolyteConverts the voltageto pH

Very cheap, accurateMust be calibratedwith a buffer solution

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pH indicatorsIndicators are dyes that can be added thatwill change color in the presence of an acidor base.

S ome indicators only work in a specificrange of pH

O nce the drops are added, the sample isruined

S ome dyes are natural, like radish skin or red cabbage

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ACID-BAS E REACTIONS

Titrations

ACID-BAS E REACTIONS

TitrationsH2C2O 4 (aq) + 2 Na O H(aq) --->

acid base

Na 2C 2O 4 (aq) + 2 H 2O (liq)Carry out this reaction using a TITR A TION .

Oxalic acid,Oxalic acid,

HH22CC 22OO 44

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S etup for titrating an acid with a baseS etup for titrating an acid with a base

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TitrationTitration1. Add solution from the

buret.2. Reagent (base) reacts

with compound (acid) insolution in the flask.

3. Indicator shows whenexact stoichiometricreaction has occurred.

(Acid = Base)T his is called

NEU T RALIZAT IO N.

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PROBLEM: You have 50.0 mL of 3.0M NaOH and you want 0.50 MNaOH. What do you do?

PROBLEM: You have 50.0 mL of 3.0M NaOH and you want 0.50 MNaOH. What do you do?

3 .0 M N aOH 0.50 M N aOH

H2

O

Concentrated Dilute

But how much water But how much water

do we add?do we add?

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PROBLEM: You have 50.0 mL of 3.0M NaOH and you want 0.50 MNaOH. What do you do ?

PROBLEM: You have 50.0 mL of 3.0M NaOH and you want 0.50 MNaOH. What do you do ?

How much water is added ?

T he important point is that --->

moles of NaOH in ORIGINAL solution =moles of NaOH in ORIGINAL solution =

moles of NaOH in FINAL solutionmoles of NaOH in FINAL solution

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PROBLEM: You have 50.0 mL of 3.0 M NaOHand you want 0.50 M NaOH. What do you do?PROBLEM: You have 50.0 mL of 3.0 M NaOHand you want 0.50 M NaOH. What do you do?

Amount of Na O H in original solution =

M V =

(3.0 mol/L)(0.050 L) = 0.15 mol Na O H

Amount of Na O H in final solution must also= 0.15 mol Na O H

Volume of final solution =

(0.15 mol Na O H) / (0.50 M) = 0.30 L

or 300 mL

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PROBLEM: You have 50.0 mL of 3.0 MNaOH and you want 0.50 M NaOH. What

do you do?

PROBLEM: You have 50.0 mL of 3.0 MNaOH and you want 0.50 M NaOH. What

do you do?Conclusion:

add 2 50 mL of

water to 50.0mL of 3.0 MNa O H tomake 300mL of 0.50 MNa O H.3 .0 M N aOH 0.50 M N aOH

H2 O

Concentrated Dilute

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A shortcut

M1 V1 = M 2 V2

Preparing S olutions by

Dilution

Preparing S olutions by

Dilution

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You try this dilution

problemYou have a stock bottle of hydrochloric acid, which is 12.1 M.

You need4

00 mL of 0.10 M HCl.How much of the acid and howmuch water will you need ?

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Acid/Base definitions D efinition #1: Arrhenius (traditional)

Acids ± produce H+

ions (or hydroniumions H 3O +)

Bases ± produce O H-

ions

(problem: some bases don¶t have

hydroxide ions!)

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Arrhenius acid is a substance that produces H + (H3O +) in water

Arrhenius base is a substance that produces O H- in water

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HN O 3, HCl, H 2SO 4 and HClO 4 are among theonly known strong acids .

S trong and Weak Acids/BasesS trong and Weak Acids/Bases

The strength of an acid (or base) isdetermined by the amount of ION IZATIO N .

The strength of an acid (or base) isdetermined by the amount of ION IZATIO N .

HO N ORS O N LY!HO N ORS O N LY!

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S trong and Weak Acids/BasesS trong and Weak Acids/BasesGenerally divide acids and bases into S T RO NG or Generally divide acids and bases into S T RO NG or WEAK ones.WEAK ones.

S T RO NG ACI D : HN OS T RO NG ACI D : HN O 33 (aq) + H(aq) + H 22O (l)O (l) ------>>HH33OO ++ (aq) + N O(aq) + N O 33

-- (aq)(aq)

HNOHNO 33 is about 100% dissociated in water.is about 100% dissociated in water.

HO N ORS O N LY!HO N ORS O N LY!

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Weak acidsWeak acids are much less than 100% ionized inare much less than 100% ionized inwater.water.

O ne of the best known is acetic acid = CHO ne of the best known is acetic acid = CH 33COCO 22HH

S trong and Weak Acids/BasesS trong and Weak Acids/BasesS trong and Weak Acids/BasesS trong and Weak Acids/BasesHO N ORS O N LY!HO N ORS O N LY!

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S trong Base:S trong Base: 100% dissociated in water.100% dissociated in water.Na O H (aq)Na O H (aq) ------> Na> Na ++ (aq) + O H(aq) + O H-- (aq)(aq)

S trong and Weak Acids/BasesS trong and Weak Acids/Bases

Other common strongOther common strongbases include KOH andbases include KOH andCa(OH)Ca(OH)

22..

CaO (lime) + HCaO (lime) + H 22OO ---->>

Ca(OH)Ca(OH) 22 (slaked lime)(slaked lime)CaOCaO

HO N ORS O N LY!HO N ORS O N LY!

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Weak base: less than 100% ionized in water Weak base: less than 100% ionized in water O ne of the best known weak bases is ammoniaO ne of the best known weak bases is ammoniaNHNH33 (aq) + H(aq) + H 22O (l)O (l) NHNH 44

++ (aq) + O H(aq) + O H-- (aq)(aq)

S trong and Weak Acids/BasesS trong and Weak Acids/BasesS trong and Weak Acids/BasesS trong and Weak Acids/BasesHO N ORS O N LY!HO N ORS O N LY!

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Weak BasesWeak BasesHO N ORS O N LY!HO N ORS O N LY!

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Acid/Base Definitions

D efinition #2: Brønsted ± Lowry

Acids ± proton donor

Bases ± proton acceptor

A ³proton´ is really just ahydrogen atom that has lost it¶selectron!

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A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor

acid conjugatebase

base conjugateacid

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ACID-BAS E THEORIESACID-BAS E THEORIES

T he Brønsted definition means NH 3 is aBAS E in water ² and water is itself an

ACI D

BaseAcidAcidBase

NH 4+ + OH -NH 3 + H 2 O

BaseAcidAcidBase

NH 4+ + OH -NH 3 + H 2 O

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Conjugate Pairs

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Acids & Base Definitions

Lewis acid - a substancethat accepts an

electron pair

Lewis baseLewis base - - aasubstance thatsubstance thatdonates an electrondonates an electronpair pair

Definition #3 ± LewisDefinition #3 ± Lewis

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Formation of hydronium ion is alsoan excellent example .

Lewis Acids & Bases

Electron pair of the new OElectron pair of the new O- -H bondH bondoriginates on the Lewis base .originates on the Lewis base .

HH

H

B A S E

O ³ H

O ³ H

H +

A C I D

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Lewis Acid/Base Reaction

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Lewis Acid-Base Interactions

in BiologyT he heme group inhemoglobin caninteract with O 2 and

CO .T he Fe ion inhemoglobin is aLewis acid

O 2 and C O can act asLewis bases

Heme group

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Calculating the pH

pH = - log [H+](Remember that the [ ] mean Molarity)

Example: If [H+

] = 1 X 10-10

pH = - log 1 X 10 -10

pH = - (- 10)pH = 10

Example: If [H+

] = 1.8 X 10-5

pH = - log 1.8 X 10 -5

pH = - (- 4 .7 4 )pH = 4 .7 4

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Try These!

Find the pH of these:1) A 0.15 M solution

of Hydrochloricacid

2) A 3.00 X 10 -7 M

solution of Nitricacid

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pH calculations ² S olving for

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pH calculations ² S olving forH+

A solution has a pH of 8.5.What is the Molarity of hydrogenions in the solution ?

pH =pH = -- log [Hlog [H ++]]8 .5 =8 .5 = -- log [Hlog [H ++]]

--8 .5 = log [H8 .5 = log [H ++]]

A ntilogA ntilog - -8 .5 = antilog (log [H8 .5 = antilog (log [H ++])])1010 --8 .58 .5 = [H= [H ++]]

3 .16 X 1 03 .16 X 1 0 --99 = [H= [H ++]]

pH =pH = -- log [Hlog [H ++]]8 .5 =8 .5 = -- log [Hlog [H ++]]

--8 .5 = log [H8 .5 = log [H ++]]

A ntilogA ntilog - -8 .5 = antilog (log [H8 .5 = antilog (log [H ++])])1010 --8 .58 .5 = [H= [H ++]]

3 .16 X 1 03 .16 X 1 0 --99 = [H= [H ++]]

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pOH S ince acids and bases are

opposites, pH and p O H areopposites!pO H does not really exist, but it isuseful for changing bases to pH.pO H looks at the perspective of abase

pO H = - log [ O H-]

S ince pH and p O H are on oppositeends,pH + p O H = 1 4

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SIPENM ARU 1988

11. S uatu larutan mempunyai pH 1. berapagaram N aOH padat ( M r = 40) harusditambahkan pada satu liter larutan ini, untukmenaikkan pHnya menjadi 3?...gram

A.0,04 C. 0,4 E . 3,96

B. 4,0 D. 7,96

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PEM BAHAS AN

J AWAB : E

Reaksi asam dan basa dapat diseerhnakan sebagai :

H+ + OH - H2O

pH awal 1, berarti H + awal = 0 ,1M

pH akhir 3, berarti H + akhir = 0 ,00 1M

[H+] yang bereaksi = 0 ,1 ± 0 ,00 1 = 0 ,0 99M

[OH -] yang diperlukan = 0 ,0 99M( koefisien sama)

N aOH yang harus ditambahkan 0 ,0 99mol (dalam volum1liter)

Maka; 0 ,0 99 x 4 0 = 3,96 gram

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SIPENM ARU 1987

13. Kedalam 1 liter larutan asam asetat 0,2 M dimasukkan beberapa gram N aOH padat,hingga pH lartan menjadi 4. bila perubahanvolum larutan diabaikan, serta menggunakantetapan ionisasi asam asetat Ka = 2x 10 -5.

M aka jumlah N aOH yang dimasukkanadalah««gram( N a = 23, O = 16, H = 1)

A. 1,33 C. 2,33 E . 3,33B. 2,0 D. 3,0

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PEM BAHAS AN

J AWAB : A

CH3COOH + N aOH CH3COON a + H 2OM : 0,2 X - -

R : X X X XS : 0,2-x - X X

[H+] =Ka x mol asam

mol garam

10 -4 = 2x10 -5 x 0,2-X

X

X = 0,033 mol

Garam N aOH = 0,033 x 40 = 1,33

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SPM B 2005 Regional I , II , III

17. Berapakah pH larutan yang diperoleh dengan

mencampurkan 50ml H N O3 0,2 M dan 50 ml KOH 0,4M ?

A.2 C.. 7 E . 13

B. 5 D. 10

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PEM BAHAS AN

J AWAB : E

HN O3 = 50 x 0,2 = 10 mmol

KOH = 50 x 0,4 = 20 mmolS isa KOh = 20 ² 10 = 10 mmol

[OH- ] = b x M b = 1 x 10 = 10 -1

100

pOH = 1

pH = 13

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PP 1981

19. J ika dari zat-zat dibawah ini dibuat larutan (dalam

air) dengan konsentrasi 1 molar, larutan manakah yangmempunyai pH paling tinggi?

A. N aHS O4 C. HCl E .CH4COON a

B. N aF D. N H4Cl

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PEM BAHAS AN

J AWAB : B dan E

N aHS O4 adalah garam asam, dalam air langsungmenghasilkan ion H +

N aF adalah garam yang bersifat basa, berasal dariasam lemah HF dan basa kuat N aOH.

HCl adalah asam kuatN H4Cl adalah garam yang bersifat basa, berasal dari

asam kuat HCl dan N H4OH

CH3COON a adalah garam yang bersifat basa,bersasal dari asam lemah CH 3COOH dan basa kuatN aOH

Larutan dengan pH tinggi adalah larutan basa

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SIPENM ARU 1981

20. S atu ml larutan N aOH 1 M ditambahkan ke dalam 1

liter air, maka larutan ini akan mempunyai pH kira-kira«

A.3

B.5

C.7

D.9E .11

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PEM BAHAS AN

J AWAB : E

Proses ini adalah proses pengenceran, volum N aOH dari1ml berubah menjadi 1001 ml.

V1 x M 1 = V2 x M 21 x 1 = 1001 x X

X = 1

1001X = ~ 10 -3

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SIPENM ARU 1980

21. Tetapan hasil kali kelarutan magnesium

hidroksida dalah 2 x10-11

. J ika pH dari suatu M gCl2dengan konsentrasi 0,002 molar dinaikkan maka akanmulai trejadi endapan pada pH«.

A.8B.9

C.10

D.11E . 12

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