Download Acids and bases

Pages 21, 22, 23
Properties
Acids
Bases
Taste
Sour
Bitter
Feel
Slippery
Conductivity
Yes
Yes
Reacts with metals
(yes or no?) what
products?
Yes- forms H2 gas
No
pH range
0-7
7-14
 Hydrogen
-
+ a polyatomic?
Polyatomic ends in –ate the acid ends in –ic
Polyatomic ends in –ic the acid ends in –ous
 Hydrogen
-
+ a single element?
Hydro______ ic acid
a) HCl – Hydrochloric acid
b) HF – Hydrofluroic acid
c) HNO2- Nitrous acid
d) H2SO4 – Sulfuric acid
 If
the acids ends in an “ic” then the
polyatomic ends in the –ate form
 If the acids ends in “ous” then the
polyatomic is in the – ite form
If the acid starts in Hydro- then the formula is
Hydrogen followed by the element ending in –
ide
Hydrochloric acid= HCl
a) nitric acid - HNO3
b) acetic acid – HC2H3O2
c) hydrobromic acid - HBr
d) sulfurous acid – H2SO3
 Following
ionic naming rules
Name the cation regularly, the
polyatomic uses its regular name.
a)KOH - Potassium hydroxide
b) Ba(OH)2- Barium hydroxide
c) LiOH- Lithium hydroxide
d) NH3 – ammonia
 Use
ionic rules
Write the symbols.
Identify the charges- criss cross to
make subscripts.
 a)
Sodium hydroxide NaOH
 b)
Beryllium hydroxide Be(OH)2
 c)
Calcium hydroxide Ca(OH)2
 d)
Cesium hydroxide CsOH
 Arrhenius
acid- contains an H+
 Bronsted-
Lowry acid- Donates an H+
 Arrhenius
base- contains an OH-
 Bronsted-
Lowry base- Accepts an H+
 HNO3
ACID
 N2H4
ACID
 N2H4
BASE
+ H2O  H3O+ + NO3BASE
ACID
BASE
+ H2O  N2H3- + H3O+
BASE
BASE
ACID
+ HCl  N2H5+ + ClACID
ACID
BASE
 Neutralization
reactions:
ACID+ BASE  SALT + WATER
KOH + H2CO3 
KOH + H2CO3  H2O + K2CO3
Don’t forget to BALANCE!
2KOH + H2CO3  2H2O + K2CO3
HBr + Al(OH)3 
HBr + Al(OH)3  H2O + AlBr3
3HBr + Al(OH)3  3H2O + AlBr3
[H+]
pH
[OH-]
pOH
Acid,
base,
neutral?
1.
2.3x10-5 M
4.64
4.3x10-10 M 9.36
Acid
2.
5.0x10-8 M
7.3
2.0x10-7 M
6.7
Base
3.
1.2x10-11 M
10.9
8.1x10-5 M
3.1
Base
 Titration-
method for determining
concentration of a solution by reacting a
known volume of solution with a
solution of known concentration
 Equivalence point- equal amounts of OH& H+ ions
Pages 24
94. Chemical and Nuclear Reaction
Radiation Type
Alpha
Beta
Gamma
+2
-1
0
Description
Helium nucleus
electron
EMR
Symbol
4
Charge
Mass
Penetrating Power
Shielding Needed
2a
or 42He
0
-1b
or 0-1e
g
4 amu
low
1/1840 amu
medium
0
high
Paper,
cloth,
skin, etc.
Aluminum
foil
Lead
Beta decay = electron
 191 is gold’s Atomic mass, which goes on top.
 Look up Gold (Au) on your periodic table to find
the atomic number which goes on the bottom.
Gold’s atomic number is 79.

 19179Au
 19179Au

0
-1e
+_____
 0-1e + 19180Hg
alpha decay = helium particle
 90 is Rubidium’s Atomic mass, which goes on
top.
 Look up Rubidium (RB) on your periodic table to
find the atomic number which goes on the
bottom. Rubidium’s atomic number is 37.

 9037Rb
 42He +_____
 9037Rb
 42He + 8635Br
A.
4
4
14 N  _____ + 1 H
He
+
2
7
1
14 N  17 C +
He
+
2
7
6
B.
102
44Ru
102
44Ru
1
1H
+ 42He 10n +_____
+ 42He 10n + 10546Pd
Page 25
 Potential
energy- stored energy due to
position
 Kinetic
energy- energy of motion
*remember that Temperature is a measure
of the average kinetic energy
 Heat-
(Q) = the process of flowing
from warmer to colder temperature.
 Temperature-
measure of the
average kinetic energy (KE) in a
sample
 Specific
heat (c) is the amount of energy
required to raise the temperature of a 1
gram sample by 1 degree
 High specific heat means that the substance
warms and cools slowly. It resists changes in
temperature.
 Low specific heat means that the substance
warms and cools quickly.
 SI
Unit = J/(g·°C)
a. Conduction- heat is transferred by touch.
Ex: heating a pan on the stove
b. Convection- heat is transferred through
liquids or gases.
Ex: Cooking in an oven
c. Radiation- heat from the sun

a. -200 kJ exothermic

b. 32 kJ

c. 653.8368 kJ endothermic
endothermic
Exothermic reactions release energy so they lose
energy (negative sign)
Endothermic reactions absorb heat so they have
energy added (positive sign)
FORMULA: Q=mc∆T
Q= heat(J); m= mass(g); c=specific heat J/(g·°C);
∆T = change in temperature (Final– initial)
Q= (1.05)(.450)(63.5)
Q= 30.0 J
∆T = 88.5-25 = 63.5
The reaction is endothermic because it
absorbs heat.
Balance the equation:
1CH4(g) + 2O2(g) 1CO2(g) +2H2O(l)
Hrxn = -890.2 kJ/mole
52.4g
1 mol
16.043 g
3.27 mol
=3.27 moles
-890.2kJ = -2910 kJ
1 mole
Page 26, 27, 28
1. Pressure (P)- atm, torr, kpa, mmHg, psi
2. Volume- (V) Liters
3. Temperature (T)- Kelvin
4. Amount- (n) moles
R = gas constant!
1. Gases consist of molecules whose separation is
much larger than the size of the molecules
themselves.
2. Particles in a gas move in straight line paths and
random directions.
3. Particles in a gas collide frequently with the
sides of the container and less frequently with
each other. All collisions are elastic (no energy
is gained or lost as a result of the collisions).
4. Particles in a gas do not attract or repel one
another. They do not sense any intermolecular
forces.
 STP
= 0 0C and 1 atm
 Temperature in Kelvin = 273K
STP can be found on the STAAR
chemistry reference chart under the
constants and conversions section.
LAW
BOYLE’S
CHARLES’S
GAY LUSSAC’S
AVOGADRO’S
COMBINED
IDEAL
INDEP/DEP
VARIABLES
CONTROL
VARIABLES
MATH
RELATIONSHIP
V, P
T, n
↑V, ↓ P
inverse, indirect
T, V
P, n
↑T ↑V
direct
P, T
V, n
V, n
↑T ↑P
direct
P, T
↑n ↑V
direct
V, P, T,
NA
P,V,n, R, T
R
NA
FORMULA
V1P1=V2P2
V1 = V2
T1
T2
P1 = P2
T1
T2
V1 = V2
n1
n2
P1V1 = P2V2
T1
T2
PV=nRT
 STP=
Standard temperature and pressure
P= 1 atm, T= 0 oC or 273K
PV=nRT
(1atm)(V)= (1.02moles)(.0821)(273)
V= 22.8L
 STP=
Standard temperature and pressure
P= 1 atm, T= 0 oC or 273K
PV=nRT
(1atm)(1.5L)=(n)(.0821)(273)
1.5= 22.4n
n= .067 moles
.067 moles
4.003g
1 mole
Molar mass of
helium
= .27 grams
 PV=nRT
(.988atm)(1.20L)=(.0470)(.0821)(T)
1.1856= .00385T
308K=T
Convert grams to moles to plug into the ideal
gas law equation!
3.58g
1 mol
=.177 mol
20.180g
PV=nRT
(.900atm)(V)=(.177mol)(.0821)(287)
V= 4.63L
 Comparing
Volume and Pressure – Boyles’ law
V1P1=V2P2
(6L)(101kPa)=(V2)(91kPa)
606=(V2)(91kPa)
6.7L=(V2)
 Comparing
Volume and pressure- Boyles’ law
V1P1=V2P2
(2.25L)(164kPa)=(1.50L)(P2)
369= (1.50L)(P2)
246 kPa=(P2)
 Comparing
Temperature, volume and
Pressure – Combined gas law
P1V1 = P2V2
T1
T2
(10.5)(.948) = (25.0)(P2)
500
618
 .019908= (25.0)(P2)
618
 12.3= (25.0)(P2)
 P2 = .49 atm

 Comparing
volume and temperature- Charles
(7.36L) = (V2)
(323K)
(173K)
(V2) = 3.94L
 Dalton’s
law of partial pressure
*Be sure that all of the pressure values have
the same units.
PTOT = P1+ P2+P3
PTOT = 1.2atm +.75atm +.41atm
PTOT = 2.36 atm
Dalton’s law of partial pressure
*Be sure that all of the pressure values have the
same units.
Convert kPA to atm 101kPa=1atm:

199 kPa
1 atm
101kPa
PTOT = P1+ P2+P3
1.97= .59+.65 +P3
.73atm =P3
= 1.97atm