chapter 8 Flashcards
define solution
a homogeneous mixture of two or more substances
define solvent and solute
solvent- substance present in the largest quantity (only 1)
solute- substance(s) present in lesser quantity
define aqueous solution
where water is the solvent
define concentration
amount of solute dissolved in a given amount of solvent or solution
define net ionic equation
shows only the atoms/molecules involved in the rxn
define full ionic equation
shows whats happening in solution
define molecular (chemical) equation
a shorthand. not separated out
define solubility
how much of a particular solute can dissolve in a given volume of solution.
define saturated
maximum amount of solute dissolved (per volume) at a given temperature/pressure
define unsaturated
sub-maximum amount of a solute dissolved(per volume)at a given temperature/pressure
define supersaturated
over maximum but stays soluble until precipitation is seeded
describe molarity( part of concentration)
moles solute/liters of solution (mol/L). “molar solution” ex: 125g NaCl in 2.5L of solution. g NaCl divided by molar mass. get answer divide by liters and get answer. this says for every 1L there is x mols
ex of M concentration: “how many moles of NaOH are in 10.0 mL of a 5.0M NaOH solution?”
(10mL)(1L/1000mL)(5.0 mol/1L)=0.050 moles
ex of M concentration: “how many liters do you need to make a 0.2 M NaOH solution with 2.2 g NaOH?”
(2.2 g NaOH)(1 mol/40.00 g NaOH)(1L/0.2 mol)=0.275 L
formula for molarity and dilutions
mol/L
(M1)(V1) —–> (M2)(V2)
stock. working
concentration concentration
how to tell if its a molarity dilution problem?
first M given is more than what you want out of it
will say “by diluting’
set up formula: “make 50 mL of a 0.05 M working solution from a 2.0 M stock solution of___”
(2.0M)(V1)/(0.05M)(0.05L)
define electrolyte and example:
ions in a solution, formed from solutes that dissociate into ions. (carry an electric current).
H2O
ex: NaCl(s) —–> Na+(aq) + Cl-(aq
1 mol. 2 mol
ONLY IONIC COMPOUND BREAK DOWN
define nonelectrolyte and example:
no ions in solution, dissolves but doesnt dissociate. (stays covalently bonded molecule)
H2O
glucose(s) ——–> glucose (aq)
1 mol. 1 mol
define strong electrolyte
dissociates virtually 100%
ex” NaCl thing
define weak electrolyte
only partially dissociates
ex: CH3COOH(aq) <====> H+(aq) + CH3COO-
EQUILIBRIUM
define acid and ex
donates H+ to solution(in water)
ex: HCl(aq) —> H+(aq) + Cl-(aq)
define base and ex
accepts H+ from solution(inH2O)
ex: NH3(aq) +H2O –> NH4+ + OH-(aq)
water acts as acid
whats acids and bases theory called
brønsted-lowry theory
describe hydronium ion when acid and bases
H+ in solution are combined with H2O to form H3O+
ex: from acid- H+(aq) + H2O(l) —> H3O+ (aq)
describe hydrochloric acid base process
H2O(l) + HCl(aq) —> Cl-(aq) + H3O+(aq)
WATER ACTS AS BASE
define amphoteric
can behave as acid and base (WATER) (H2CO3)
define monoprotic acids and diprotic acids and how
mono- loses 1H+
di- loses 2 H+
poly exists.
DI AND POLY GO THRPUGH RXN MULTIPLE TIMES TILL H IS OUT
what is the heart of all acid/base chemistry
H+(aq) + OH-(aq) —> H2O(l)
how to solve neutralization rxn: “you spill 20.0 mL of 3.5 M HCl in lab. what volume of 5M NaOH is required to neutralize the acid?”
set up ionic equation and get net(H+ + OH- —> H2O(l) ). use spill given(20mL) convert to L multiple by (3.5 mol HCl/1L) then do molar ratio from net (1 molH+/ 1mol HCl) and get 0.07 mol H+
take that answer as given then do molar ratio pf net (1 mol OH-/ 1 mop H+) then do another molar ratio ( 1 mol NaOH/ 1 mol OH-) and then do (1L/ 5 mol NaOH) and get 0.014 L 5M NaOH or 14 mL
define oxidation
lose of electrons
gain O lose H
define reduction
gain of electrons
lose O gain H
ex of redox reaction (reduction and oxidation happen simultaneously)
ANY ELEMENT YOU FIND IN NATURE = zero neutral
Mg°(s) + Cl°2(g)—>Mg²+(aq) + 2Cl-(aq)
Mg lost 2e-(ox) Cl gain 2e-(re)
break into 1/2 rxns.
Mg: Mg°(s)—> Mg²+(aq) + 2e-
Cl: Cl°2(g) + 2e- —> 2Cl-(aq)
do 1/2 redox rxns where you have to multiple so e- are same : Al°(s)+ O°2(g) –> Al2O3(s)
ox1/2:(Al°—>Al3+ +3e-) x 4
red1/2:(O°2+ 4e- –> 2O2-) x3
4Al°(s)—>4Al3+(aq) + 12e-
3O°2 + 12e- –> 6O(2-)(aq)
4Al°(s) + 3O°2 –> 4Al(3+) +6O(2-)
2Al2O0
