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The Chemistry of Acids and Bases. Acid and Bases. Acid and Bases. Acid and Bases. Acids. Have a sour taste. Vinegar is a solution of acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon - PowerPoint PPT Presentation
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11
The Chemistry of Acids and BasesThe Chemistry of Acids and Bases
22
Acid and BasesAcid and Bases
33
Acid and BasesAcid and Bases
44
Acid and BasesAcid and Bases
55Acids
Have a sour taste. Vinegar is a solution of acetic acid. Citrus fruits contain citric acid.
React with certain metals to produce hydrogen gas.
React with carbonates and bicarbonates to produce carbon dioxide gas
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Bases
66
Some Properties of Acids
Produce H+ (as H3O+) 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
pH is less than 7
Turns blue litmus paper to red “Blue to Red A-CID”
77
Anion Ending Acid Name
-ide hydro-(stem)-ic acid
-ate (stem)-ic acid
-ite (stem)-ous acid
Acid Nomenclature Review
Binary
Ternary
An easy way to remember which goes with which…
“In the cafeteria, you ATE something ICky”
88Acid Nomenclature Flowchart
99
carbonic acid
• HBr (aq)
• H2CO3
• H2SO3
hydrobromic acid
sulfurous acid
Acid Nomenclature Review
1010
Name ‘Em!
• HI (aq) _________________________
• HCl (aq) _________________________
• H2SO3 _________________________
• HNO3 _________________________
• HIO4 _________________________
1111
Some Properties of Bases
Produce OH- ions in water
Taste bitter, chalky
Are electrolytes
Feel soapy, slippery
React with acids to form salts and water
pH greater than 7
Turns red litmus paper to blue “Basic Blue”
1212
Some Common BasesSome Common Bases
NaOH sodium hydroxide lye
KOH potassium hydroxide liquid soap
Ba(OH)2 barium hydroxide stabilizer for plastics
Mg(OH)2 magnesium hydroxide “MOM” Milk of magnesia
Al(OH)3 aluminum hydroxide Maalox (antacid)
1313
Acid/Base definitions
• Definition #1: Arrhenius (traditional)
Acids – produce H+ ions (or hydronium ions H3O+)
Bases – produce OH- ions
(Problem: Some bases don’t have hydroxide ions!)
1414Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH‒ in water
1515
Acid/Base Definitions
• Definition #2: Brønsted – Lowry
Acids – proton donor
Bases – proton acceptor
A “proton” is really just a hydrogen atom that has lost it’s electron!
1616
A Brønsted-Lowry acid is a proton donorA Brønsted-Lowry base is a proton acceptor
acidconjugate
basebase conjugate
acid
1717
ACID-BASE THEORIESACID-BASE THEORIES
The Brønsted definition means NH3 is a BASEBASE in water — and water is itself an ACIDACID
BaseAcidAcidBaseNH4
+ + OH-NH3 + H2OBaseAcidAcidBase
NH4+ + OH-NH3 + H2O
1818
Conjugate PairsConjugate Pairs
1919
Learning Check!
Label the acid, base, conjugate acid, and conjugate base in each reaction:
HCl + OH ‒ Cl ‒ + H2OAcid Base Conj. Conj.
Base Acid
HCl + OH ‒ Cl ‒ + H2OAcid Base Conj. Conj.
Base Acid
H2O + H2SO4 HSO4‒ + H3O
+
Base Acid Conj. Conj. Acid Base
H2O + H2SO4 HSO4‒ + H3O
+
Base Acid Conj. Conj. Acid Base
2020Acids & Base Acids & Base DefinitionsDefinitions
Definition #3 - Lewis
Lewis acid - a substance that accepts an electron pair
Lewis base - a substance that donates an electron pair
2121
Formation of hydronium ion H3O+ is also an excellent example.
Lewis Acids & Bases
•Electron pair of the new O-H bond originates on the Lewis base.
HH
H
BASE
••••••
O—HO—H
H+
ACID
2222
Lewis Acid/Base Reaction
2323
Lewis Acid-Base Interactions in Biology
• The heme group in hemoglobin can interact with O2 and CO.
• The Fe ion in hemoglobin is a Lewis acid
• O2 and CO can act as Lewis bases
Heme group
2424The pH scalepH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H+ (or OH-) ion.
Under 7 = acid7 = neutralOver 7 = base
2525
pH of Common pH of Common SubstancesSubstances
2626Calculating 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 = - (- 4.74)
pH = 4.74
2727
Try These!
Find the pH of these:
1) A 0.15 M solution of Hydrochloric acid
2) A 3.00 X 10-7 M solution of Nitric acid
‒ Log (0.15) = 0.82
‒ Log (3.00 x 10-7) = 6.5
2828pH calculations – Solving for pH calculations – Solving for H+H+pH calculations – Solving for pH calculations – Solving for H+H+
If the pH of Coke is 3.12, [H+] = ???
Because pH = - log [H+] then
- pH = log [H+]
Take antilog (10x) of both sides and get
10-pH = [H+][H+] = 10-3.12 = 7.6 x 10-4 M *** to find antilog on your calculator, look for “Shift” or “2nd
function” and then the log button
2929pH calculations – Solving for pH calculations – Solving for H+H+
• A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution?
pH = - log [H+]
8.5 = - log [H+]
-8.5 = log [H+]
Antilog -8.5 = antilog (log [H+])
10-8.5 = [H+]
3.16 X 10-9 = [H+]
pH = - log [H+]
8.5 = - log [H+]
-8.5 = log [H+]
Antilog -8.5 = antilog (log [H+])
10-8.5 = [H+]
3.16 X 10-9 = [H+]
3030
HH22O can function as both an ACID and a BASE.O can function as both an ACID and a BASE.
In pure water there can be In pure water there can be AUTOIONIZATIONAUTOIONIZATION
Equilibrium constant for water = KEquilibrium constant for water = Kww
KKww = [H = [H33OO++] [OH] [OH--] = ] = 1.00 x 101.00 x 10-14-14 at 25 at 25 ooCC
More About Water
3131
More About WaterMore About Water
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
In a neutral solution [H3O+] = [OH-]
so Kw = [H3O+]2 = [OH-]2
and so [H3O+] = [OH-] = 1.00 x 10-7 M
OH-
H3O+
OH-
H3O+
AutoionizationAutoionization
3232pOHpOH
• Since acids and bases are opposites, pH and pOH are opposites!
• pOH does not really exist, but it is useful for changing bases to pH.
• pOH looks at the perspective of a base
pOH = - log [OH-]Since pH and pOH are on opposite
ends,pH + pOH = 14
3333
pHpH [H+][H+] [OH-][OH-] pOHpOH
3434
[H[H33OO++], [OH], [OH--] and pH] and pHWhat is the pH of the What is the pH of the
0.0010 M NaOH solution? 0.0010 M NaOH solution?
[OH-] = 0.0010 (or 1.0 X 10[OH-] = 0.0010 (or 1.0 X 10-3-3 M) M)
pOH = - log 0.0010pOH = - log 0.0010
pOH = 3pOH = 3
pH = 14 – 3 = 11pH = 14 – 3 = 11
OR KOR Kww = [H = [H33OO++] [OH] [OH--]]
[H[H3OO++] = 1.0 x 10] = 1.0 x 10-11-11 M M
pH = - log (1.0 x 10pH = - log (1.0 x 10-11-11) = 11.00) = 11.00
3535
The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater?
The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?
3636Calculating [H3O+], pH, [OH-], and pOH
Problem 1: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H3O+], pH, [OH-], and pOH of the two solutions at 25°C.
Problem 2: What is the [H3O+], [OH-], and pOH of a solution with pH = 3.67? Is this an acid, base, or neutral?
Problem 3: Problem #2 with pH = 8.05?
3737
HNO3, HCl, H2SO4 and HClO4 are among the only known strong acids.
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
The strength of an acid (or base) is determined by the amount of IONIZATION.
The strength of an acid (or base) is determined by the amount of IONIZATION.
3838
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
• Generally divide acids and bases into STRONG or Generally divide acids and bases into STRONG or WEAK ones.WEAK ones.
STRONG ACID:STRONG ACID: HNOHNO3 3 (aq) + H(aq) + H22O (l) --->O (l) --->
HH33OO+ + (aq) + NO(aq) + NO33- - (aq)(aq)
HNOHNO33 is about 100% dissociated in water. is about 100% dissociated in water.
3939
• Weak acids are much less than 100% ionized in
water.
One of the best known is acetic acid = CH3CO2H
Strong and Weak Acids/Bases
Strong and Weak Acids/Bases
4040
• Strong Base:Strong Base: 100% dissociated in 100% dissociated in water.water.
NaOH (aq) ---> NaNaOH (aq) ---> Na+ + (aq) + OH(aq) + OH- - (aq)(aq)
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
Other common strong Other common strong bases include KOH and bases include KOH and Ca(OH)Ca(OH)22..
CaO (lime) + HCaO (lime) + H22O -->O -->
Ca(OH)Ca(OH)22 (slaked lime) (slaked lime)CaOCaO
4141
• Weak base: less than 100% ionized in water
One of the best known weak bases is ammonia
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
4242
Weak Bases
4343
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Consider acetic acid, HCConsider acetic acid, HC22HH33OO22 (HOAc) (HOAc)
HCHC22HH33OO22 + H + H22O O H H33OO++ + C + C22HH33OO22 --
AcidAcid Conj. base Conj. base
Ka [H3O+][OAc- ]
[HOAc] 1.8 x 10-5Ka
[H3O+][OAc- ][HOAc]
1.8 x 10-5
(K is designated K(K is designated Kaa for ACID) for ACID)
K gives the ratio of ions (split up) to molecules K gives the ratio of ions (split up) to molecules
(don’t split up)(don’t split up)
HONORS ONLY!HONORS ONLY!
4444Ionization Constants for Ionization Constants for Acids/Bases Acids/Bases
AcidsAcids ConjugateConjugateBasesBases
Increase strength
Increase strength
HONORS ONLY!HONORS ONLY!
4545
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
Weak acid has KWeak acid has Kaa < 1 < 1
Leads to small [HLeads to small [H33OO++] and a pH of 2 - 7] and a pH of 2 - 7
HONORS ONLY!HONORS ONLY!
4646
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Weak base has KWeak base has Kbb < 1 < 1
Leads to small [OHLeads to small [OH--] and a pH of 12 - 7] and a pH of 12 - 7
HONORS ONLY!HONORS ONLY!
4747
Relation Relation
of Kof Kaa, K, Kbb, ,
[H[H33OO++] ]
and pHand pH
HONORS ONLY!HONORS ONLY!
4848Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, OAc, OAc--, ,
and the pH.and the pH.
Step 1.Step 1. Define equilibrium concs. in ICE Define equilibrium concs. in ICE
table.table.
[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]
initialinitial
changechange
equilibequilib
1.001.00 00 001.001.00 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
1.00-x1.00-x xx xx1.00-x1.00-x xx xx
HONORS ONLY!HONORS ONLY!
4949Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Step 2.Step 2. Write KWrite Kaa expression expression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - xKa 1.8 x 10-5 =
[H3O+][OAc- ][HOAc]
x2
1.00 - x
This is a quadratic. Solve using quadratic This is a quadratic. Solve using quadratic formula.formula.
or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok) or smaller is ok)or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok) or smaller is ok)
HONORS ONLY!HONORS ONLY!
5050Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Step 3.Step 3. Solve KSolve Kaa expression expression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - xKa 1.8 x 10-5 =
[H3O+][OAc- ][HOAc]
x2
1.00 - x
First assume x is very small because First assume x is very small because KKaa is so small. is so small.
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
Now we can more easily solve this Now we can more easily solve this approximate expression.approximate expression.
HONORS ONLY!HONORS ONLY!
5151
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve KSolve Kaa approximateapproximate expressionexpression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
x =x = [[HH33OO++] = [] = [OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M
pH = - log [pH = - log [HH33OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) =) = 2.372.37
HONORS ONLY!HONORS ONLY!
5252Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Calculate the pH of a 0.0010 M solution of Calculate the pH of a 0.0010 M solution of formic acid, HCOformic acid, HCO22H.H.
HCOHCO22H + HH + H22O O HCO HCO22-- + H + H33OO++
KKaa = 1.8 x 10 = 1.8 x 10-4-4
Approximate solutionApproximate solution
[H[H33OO++] = 4.2 x 10] = 4.2 x 10-4-4 M, M, pH = 3.37pH = 3.37
Exact SolutionExact Solution
[H[H33OO++] = [HCO] = [HCO22--] = 3.4 x 10] = 3.4 x 10-4-4 M M
[HCO[HCO22H] = 0.0010 - 3.4 x 10H] = 0.0010 - 3.4 x 10-4-4 = 0.0007 M = 0.0007 M
pH = 3.47 pH = 3.47
HONORS ONLY!HONORS ONLY!
5353Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 1.Step 1. Define equilibrium concs. in ICE tableDefine equilibrium concs. in ICE table
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
0.0100.010 00 000.0100.010 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx
HONORS ONLY!HONORS ONLY!
5454Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 1.Step 1. Define equilibrium concs. in ICE tableDefine equilibrium concs. in ICE table
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
0.0100.010 00 000.0100.010 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx
HONORS ONLY!HONORS ONLY!
5555Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression
Kb 1.8 x 10-5 = [NH4
+][OH- ][NH3 ]
= x2
0.010 - xKb 1.8 x 10-5 =
[NH4+][OH- ]
[NH3 ] =
x2
0.010 - x
Assume x is small, soAssume x is small, so x = [OHx = [OH--] = [NH] = [NH44
++] = 4.2 x 10] = 4.2 x 10-4-4 M M
and [NHand [NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 ≈ 0.010 M ≈ 0.010 M
The approximation is validThe approximation is valid !!
HONORS ONLY!HONORS ONLY!
5656Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 3.Step 3. Calculate pHCalculate pH
[OH[OH--] = 4.2 x 10] = 4.2 x 10-4-4 M M
so pOH = - log [OHso pOH = - log [OH--] = 3.37] = 3.37
Because pH + pOH = 14,Because pH + pOH = 14,
pH = 10.63pH = 10.63
HONORS ONLY!HONORS ONLY!
5757
Types of Acid/Base Reactions: Types of Acid/Base Reactions: SummarySummary
HONORS ONLY!HONORS ONLY!
5858pH testing
• There are several ways to test pHThere are several ways to test pH
–Blue litmus paper (red = acid)Blue litmus paper (red = acid)
–Red litmus paper (blue = basic)Red litmus paper (blue = basic)
–pH paper (multi-colored)pH paper (multi-colored)
–pH meter (7 is neutral, <7 acid, >7 pH meter (7 is neutral, <7 acid, >7 base)base)
–Universal indicator (multi-colored)Universal indicator (multi-colored)
– Indicators like phenolphthaleinIndicators like phenolphthalein
–Natural indicators like red cabbage, Natural indicators like red cabbage, radishesradishes
5959Paper testing
• Paper tests like litmus paper and pH Paper tests like litmus paper and pH paperpaper
– Put a stirring rod into the solution Put a stirring rod into the solution and stir.and stir.
– Take the stirring rod out, and Take the stirring rod out, and place a drop of the solution from place a drop of the solution from the end of the stirring rod onto a the end of the stirring rod onto a piece of the paperpiece of the paper
– Read and record the color change. Read and record the color change. Note what the color indicates. Note what the color indicates.
– You should only use a small You should only use a small portion of the paper. You can use portion of the paper. You can use one piece of paper for several one piece of paper for several tests.tests.
6060pH paperpH paper
6161
pH meter
• Tests the voltage of the Tests the voltage of the electrolyteelectrolyte
• Converts the voltage to Converts the voltage to pHpH
• Very cheap, accurateVery cheap, accurate
• Must be calibrated with Must be calibrated with a buffer solutiona buffer solution
6262pH indicators
• Indicators are dyes that can be added that will change color in the presence of an acid or base.
• Some indicators only work in a specific range of pH
• Once the drops are added, the sample is ruined
• Some dyes are natural, like radish skin or red cabbage
6363
ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations
ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations
HH22CC22OO44(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) --->
acid baseacid base
NaNa22CC22OO44(aq) + 2 H(aq) + 2 H22O(liq)O(liq)
Carry out this reaction using a TITRATION.Carry out this reaction using a TITRATION.
Oxalic acid,Oxalic acid,
HH22CC22OO44
6464Setup for titrating an acid with a baseSetup for titrating an acid with a base
6565
TitrationTitrationTitrationTitration 1. Add solution from the 1. Add solution from the buret.buret.
2. Reagent (base) reacts 2. Reagent (base) reacts with compound (acid) in with compound (acid) in solution in the flask.solution in the flask.
3.3. Indicator shows when Indicator shows when exact stoichiometric exact stoichiometric reaction has occurred. reaction has occurred. (Acid = Base)(Acid = Base)
This is called This is called NEUTRALIZATION.NEUTRALIZATION.
6666
35.62 mL of NaOH is 35.62 mL of NaOH is
neutralized with 25.2 neutralized with 25.2
mL of 0.0998 M HCl by mL of 0.0998 M HCl by
titration to an titration to an
equivalence point. equivalence point.
What is the What is the
concentration of the concentration of the
NaOH?NaOH?
LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.
LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.
6767
PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?
PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?
Add water to the 3.0 M solution to lower Add water to the 3.0 M solution to lower its concentration to 0.50 M its concentration to 0.50 M
Dilute the solution!Dilute the solution!
6868
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?
3.0 M NaOH 0.50 M NaOH
H2O
Concentrated Dilute
But how much water But how much water do we add?do we add?
6969
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you doWhat do you do??
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you doWhat do you do??
How much water is added?How much water is added?
The important point is that …The 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
7070
PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?
Amount of NaOH in original solution = Amount of NaOH in original solution =
M • V M • V = =
(3.0 mol/L)(0.050 L) = 0.15 mol NaOH(3.0 mol/L)(0.050 L) = 0.15 mol NaOH
Amount of NaOH in final solution must also = Amount of NaOH in final solution must also = 0.15 mol NaOH0.15 mol NaOH
Volume of final solution =Volume of final solution =
(0.15 mol NaOH)(1 L/0.50 mol) (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L= 0.30 L
= = 300 mL300 mL
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PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?
Conclusion:Conclusion:
Add 250 mL Add 250 mL of water to of water to 50.0 mL of 50.0 mL of 3.0 M NaOH 3.0 M NaOH to make 300 to make 300 mL of 0.50 M mL of 0.50 M NaOH.NaOH.3.0 M NaOH 0.50 M NaOH
H2O
Concentrated Dilute
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A shortcutA shortcut
MM11 • V • V11 = M = M22 • V • V22
Preparing Solutions Preparing Solutions by Dilutionby Dilution
Preparing Solutions Preparing Solutions by Dilutionby Dilution
7373You try this dilution problem
• You have a stock bottle of hydrochloric acid, which is 12.1 M. You need 400 mL of 0.10 M HCl. How much of the acid and how much water will you need?