Acids and Bases
Roselyn Dooley, Tyler Schmidt, Kyle
Doubleday and Deondré Robinson
Properties of Acids
-Sour taste
-React with active metals
-Turns litmus paper Red
-React with bases to produce salt and water
-Conduct electricity
-1- 6.9 on pH scale
Properties of Bases
-Bitter taste
-Slippery feel
-Turns litmus paper blue
-Reacts with acids to produce salt and water
-Conduct electricity
-7.1 - 14 on pH scale
Binary and Tertiary acids
Binary acid- an acid that contains only two
different elements: hydrogen and one of the
more electronegative elements.
Tertiary acid- an acid that contains hydrogen
oxygen and one more element.
Theories of Acids and Bases
-Arrhenius
Acid- A substance that dissociates to produce
hydrogen ions in water
Base- A substance that dissociates to produce
hydroxide ions in water
-Bronsted-Lowry
Acid- Any substance that can donate H+ ions. (A
proton donor)
Base- Any substance that can accept H+ ions. (A
proton acceptor)
Theories of Acids and Bases Cont.
-Lewis
Acid- Any substance that can accept a pair of
nonbonding electrons. (electron pair acceptor)
Base- Any substance that can donate a pair of
nonbonding electrons. (electron pair donor)
Naming Acids
•
Rule #1
o
•
If the negative ion in the acid ends in "ide" you name
the acid "Hyrdo (stem) ic acid
 Ex: HCl (Chloride) would yield Hydrochloric acid
Rule # 2
o
If the negative ion in the acid ends in "ite", you name
the acid "(stem)ous acid"
 Ex: HNO2 (Nitrite) would yield Nitrous Acid
*Use only if there is an oxygen in the chemical formula
Naming Acids Cont.
•
Rule # 3
o If the negative ion ends in "ate", you name the acid
"(stem)ic" acid.
 Ex: HIO4 (Periodate) would yield Periodic acid
 Note: The stem of Sulfur is Sulfur
• Also the Stem of Phosphor is phosphor
Name that Acid!
1. HSCN
2. HClO2
3. HClO3
4. HBr
5. H2SO3
6. H3P
Answers
1. Thiocyanic acid
2. Chlorous acid
3. Chloric acid
4. Hydrobromic acid
5. Sulfurous acid
6. Hydrophosphoric acis
pH Scale
Six Strong Acids
•
•
•
•
•
•
HCl
HBr
HI
HNO3
H2SO4
HClO4
o
Everything else is considered a weak acid
Writing Acid-base reactions in
aqueous solutions
General Formulas
Strong Acids
Weak Acids
HA
H+1 + A-1
HA +H2O
H3O+1 +A-1
HCl
H+1 + Cl-1
HF + H2O
H3O+1 + F-1
Now you try
Write the acid base reactions in aqueous solutions
1. HBr
2. H2SO4
3. HCN
4. HC2H3O2
Answers
1. HBr
H+1 + Br-1
2. H2SO4
H+1 + SO4-2
3. HCN + H2O
H3O+1 + CN-1
4. HC2H3O2 + H2O
H3O+1 + C2H3O2-1
Neutralization Reactions between
Acids and Bases
Neutralization- The reaction of hydronium ions
and hydroxide ions to form water molecules
Example equationHCl(aq) + NaOH(aq)
NaCl(aq)+ H2O(l)
Note: This is basically just a double
displacement reaction
Now you try
1. HClO4(aq)+ NaOH(aq)
2. HBr(aq) + Ba(OH)2 (aq)
***You might need your pink sheet
Answers
1. HClO4(aq)+ NaOH(aq)
2. 2HBr (aq)+ Ba(OH)2 (aq)
2H2O(l)
NaClO4(aq)+ H2O(l)
BaBr2 (aq) +
Calculate Hydronium and
Hydroxide
[H+]=10-pH
ex. [H+]=10-4
[H+]=1 x 10-4 M
[OH-]=10-pOH
ex. [OH-]=10-7.5
[OH-]=3.16 x 10-8 M
1.
pH= 2.2
1.
pOH=7.8
2.
pH=3.6
2.
pOH=9.3
3.
pH=8.8
3.
pOH=5.6
4.
pOH=9
4.
pH=3
Hydronium and Hydroxide Answers
1. [H+]= .0063 M
1. [OH-]= 1.58 x 10-8 M
2. [H+]= 2.51 x 10-4 M
2. [OH-]= 5.01 x 10-10 M
3. [H+]= 1.58 x 10-9 M
3. [OH-]= 2.51 x 10-6 M
4. [H+]= 1 x 10-9 M
4. [OH-]= .001 M
Calculate pH and pOH
pH=-log[H+]
ex. pH=-log[2.33 x 10-9 M]
pOH=-log[OH-]
ex. pOH=-log[7.65 x 10-3 M]
pH=8.63
1.
[H+]=7.24 x 10-5 M
2.
[H+]=6.32 x 10-2 M
3.
[OH-]=2.26 x 10-8 M
4.
[H+]=4.54 x 10-3 M
pOH=2.12
1.
[OH-]=5.58 x 10-4 M
2.
[OH-]=3.67 x 10-8 M
3.
[OH-]=2.77 x 10-2 M
4.
[H+]=4.49 x 10-7 M
pH and pOH Answers
1. pH= 4.14
1. pOH= 3.25
2. pH= 1.2
2. pOH= 7.43
3. pH= 6.35
3. pOH= 1.56
4. pH= 2.34
4. pOH= 7.65
Titration
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.
Once the two solutions are chemically
equivalent, the solution has reached the
equivalence point.
Titration
Essentially, you add an acid of known molarity
(concentration) to a base of unknown molarity
in measured amounts to find the unknown, or
vice versa.
Once at the equivalence point, the unknown
concentration can be calculated using known
concentration and volumes.
Walkthrough Problem
500 mL of .5 M HF titrates with 635 mL NaOH.
1) Balance the equation:
HF + NaOH
NaF + H2O
2) Choose method: M1V1 = M2V2 or
conversions
3) (.5M) (.500 L) = (x M) (.635 L)
4) .5 M (.500 L)
.635 L
5) .4 M NaOH
Titration Calculations
Ex 1) 25 mL of .3M HCl reaches an
equivalence point with 75 mL of NaOH. What
is the molarity of the NaOH?
Titration Calculations
1) Balance the equation:
HCl + NaOH
NaCl + H2O
2) Because the mole ratios are equal (1:1), we
can use the formula M1V1 = M2V2
3) (.3M) (.025 L) = (x M) (.075 L)
4) .3 M (.025 L)
=xM
.075 L
5) .1 M NaOH
Titration Calculations
Ex 2) 550 mL of H2SO4 of unknown
concentration reaches an equivalence point
with 775 mL of 2.0 M NaOH. What is the
concentration of the H2SO4.
Titration Calculations
1) Balance the equation:
H2SO4 + 2NaOH
Na2SO4 + 2H2O
2) Because the mole ratios are not equal (1:2),
we must use conversion factors.
3) 2.0 M NaOH = (x) mols NaOH
.775 L NaOH
4) 1.55 mols NaOH x 1 mol H2SO4 = .755 mols H2SO4
2 mols NaOH
5) .755 mols H2SO4 = 1.4 M H2SO4
.550 L H2SO4
Questions for us
Sources
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bnw=65&zoom=1&usg=__Lx_UdDHNNq5SJQMUUATJ_Hg7nCc=&docid=8iiaXS9UV6QRhM&sa=X&ei=gF6tUfDeDtCJ0QH
Fg4DgCA&sqi=2&ved=0CF4Q9QEwBg&dur=155
http://www.elmhurst.edu/~chm/vchembook/184ph.html
http://copernicusconsulting.net/designers-are-not-researchers-the-difference-between-design-and-social-research/
http://www.thechemicalblog.co.uk/what-is-a-titration/
http://dl.clackamas.edu/ch104-04/double.htm
http://www.pc.maricopa.edu/data/GlobalFiles/file/chemistry/lee/Double%20displacement%20reaction.pdf
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PdSM:&ved=0CAUQjRw&url=http%3A%2F%2Fthepickledhedgehog.com%2F2012%2F02%2F21%2Fsciencelife%2F&ei=Rl-tUYjpM8aw0AGWpoCQAw&bvm=bv.47244034,d.dmQ&psig=AFQjCNH5PXKRxbEbTHb1ihQtytljDNj1Q&ust=1370403008833830