Acids,Bases, and Salt

Portfolio Pg. 6

Section 1: Acid-Base Theories

Arrhenius said that acids produce hydrogen ions (H+) in aqueous solutions.An example would be HCL ----> H+ + Cl-. Bases produce hydroxide ions (OH-) when dissolved in water. Ammonia (NH3) is not an Arrhenius base because no OH is produced. An example would be NaOH --->Na+ + OH-.
Bronsted-Lowry stated that acids are hydrogen- ion donors. HCl is an acid. when dissolved in water the proton is donated to water. EX: HCl+ H2O <----->H3O+ +Cl-. If an acid donates then a base must accept hydrogen- ions. Ammonia is a base because of Bronsted- Lowry.
NH3 + H2O ----->NH4+ +OH-. In this situation the NH3 would act as the acid, H2O acts as the base, NH4 is the conjugate acid, and lastly the OH is the conjugate base.
Lewis said that acid electron pairs acceptance and that bases electron pairs donate. Acids do not need hydrogen.

Section 2: Calculating pH and pOH

Going from Hydrogen to find the pH you would do pH=-log[H+]. To find the pOH going from hydrogen you would do the steps above to get your pH then go from there and do pOH=14- pH.
Going from pH to [H+] you would [H+]=10 to the -pH.
From the pOH to the [OH-] you would use this formula pOH= -log[OH-]
Strong Acids lnclude HCl, HBr, HI, HNO3, HClO3,H2SO4.
Strong Bases Include LiOH, NaOH, KOH,RbOH, CsOH, Ba(OH)2,Sr(OH)2, and Ca(OH)2
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If you have a Binary acid it only has two elements, the first element is going to be hydrogen, when you go to name it it will be along the lines of hydro____ ic acid. ex: HCl hydrochloric acid. For an Ternary acid you are going to have three elements, the first is going to be hydrogen as well, other elements part of a polyatomic ion. Ate---> ic. Ite---> ous. ous is lowest. H2SO4 would be sulfuric acid. H2SO3 would be Sulfurous acid. For bases you name them by the cation ( +1 Charge) and the the anion (-1 charge). EX: Ca(OH)2 Calcium hydroxide. A salt the first part is formed of an metal in the metal oxide,hydoxide, or carbonated your second section is when a salt and water is formed. HCl + NaOH ----> NaCl + H2O.

Section 4: Ka and Kb Expressions

HA+ H2O= H+ +A- This makes the equilibrium acid dissociation constant =Ka .
Ka= [H+][A-]
[HA]
EX; HI+ H2O----> H3O+ +I
Ka= [H3O][I-]
[HI]
Kb exoressions
Kb= [conjugate acid] x [OH-]
[conjugate base]
EX; HNO2---> H+ + NO2-
Kb= [H+][NO2-]
[HNO2]
If it takes 54 mL of 0.1 M NaOH to neutralize 125 mL of an HCl solution,
what is the concentration of the HCl?
KNOWN: .054 L NaOH
.1M NaOH
.125 L HCl
UNKNOWN:
M HCl?
NaOH+ HCl ---> NaCl +H2O
.054*.1 Mol NaOH* 1 Mol HCl= .054
1 L NaOH 1 Mol NaOh .125 L HCl
Answer : 0.0432 M HCl