Acids and bases
Download
Report
Transcript 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
2a
or 42He
0
-1b
or 0-1e
g
4 amu
1/1840 amu
0
Penetrating Power
low
medium
high
Shielding Needed
Paper,
cloth,
skin, etc.
Aluminum
foil
Lead
Mass
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 + 1 H
He
+
2
7
6
1
B.
102
44Ru
102
44Ru
+ 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