ACID MIXING INSTRUCTIONS
ALL BASED ON MILLILITERS OF CONCENTRATED
ACID PER LITER OF MIXED SOLUTION
NITRIC ACID
1 MOLAR...... 64ml/LITER
2 MOLAR...... 128ml/LITER
6 MOLAR...... 384ml/LITER
HYDROCHLORIC ACID
.1 MOLAR...... 8.6ml/LITER
.5 MOLAR...... 42.8ml/LITER
1 MOLAR...... 85.5ml/LITER
2 MOLAR...... 171ml/LITER
6 MOLAR...... 513ml/LITER
SULFURIC ACID
1 MOLAR...... 55.5ml/LITER
2 MOLAR...... 111ml/LITER
6 MOLAR...... 340ml/LITER
Information Sheet: ACIDS-BASES-SALTS
ACIDS-Taste sour
Turn Litmus pink
Corrode metals to form H2
Harmful to living tissue
Neutralize all bases
Conduct electricity
Give off H+ in aq. Solution (Arrhenius theory)
Give off protons in aq. Soln. (Bronsted-Lowry theory)
Accepts 2 e-‘s forming a covalent bond (Lewis Theory)
Monoprotic, Diprotic, Triprotic (all H+’s are not released at once)
BASES-Taste bitter
Turn litmus blue
Harmful to living tissues
Neutralize all acids
Conduct electricity
Give off OH- ions in aq. Solutions (Arrhenius)
Accept protons in aq. Solution (Bronsted-Lowry)
Donates 2 e-‘s forming covalent bond (Lewis Theory)
SALTS-Product of an ACID-BASE neutralization
Ionic compounds
Contain POSITIVE ion from base; Contain NEGATIVE ion from acid
Usually have high melting points
Conduct electrical current in aq. solutions
Molten form also conducts electricity
NEUTRALIZATION OF ACIDS AND BASES (be sure to check charges and BALANCE equation)
ACID + BASE --> WATER + SALT
HCl + NaOH --> H20 + NaCl
H+ + Cl- + Na+ + OH- --> H2O + NaCl (IONS FORMING SALT IS BASED UPON SOLUBILITY
H2SO4 + Ba(OH)2 --> 2H20 + BaSO4
EQUILIBRIUM OF [H+] and [OH-} in aqueous solution.
Because of neutralization, the amount of acid affects the amount of base present in aqueous solutions.
MORE ACID--LESS BASE : MORE BASE--LESS ACID
Kw – a mathematical relationship between [H+] and [OH-] Kw = 1E-14
Kw = [H+][OH-]
Kw = 1E-14 = [H+][OH-]
Example: if the [H+] = 2.5E-3 M, find the [OH-] of the solution.
Kw = [H+][OH-]
1E-14 = (2.5E-3)[OH-]
[OH-] = 1E-14 / 2.5E-3 = 4E-12 M
Example: if the [OH-] = 3.33E-5 M, find the [H+] of the solution
Kw = [H+][OH-]
1E-14 = [H+](3.33E-5)
[H+] = 1E-14 / 3.33-5 = 3E-10 M
pH = -(Log[H+])
N
pH Scale from 0------7------14
Acids | Bases
Strong Weak|Weak Strong
Weak Acid Dissociation: HA à H+ + A- where HA does NOT dissociate 100% but reaches equilibrium.
[H+][A-]
Ka = ______
[HA]
Acid-Base Worksheet I Kw Problems Name ______
1. Find the [OH-] in a .0023 M HCl solution. (4.35e-12)
2. Find the [H+] in a .000005 M NaOH solution. (2e-9)
3. Find the [H+] in a 8.9E-4 M Ba(OH)2 solution. (5.62e-12)
4. Find the [H+] in a 3.4e-4 M HNO3 solution. (3.4e-4)
pH Problems
pH = -(Log[H+])
N
pH Scale from 0------7------14
Acids | Bases
Strong Weal|Weak Strong
Example: If the [H+] = 2.5E-5, find the phFIND THE pH.
pH= -(LOG(2.5E-5))
= -(-4.6020)
= 4.6020
Example: if the pH=2.3445, find the [H+].
[H+]= -(ANTILOG(-2.3445))
[H+]= 4.5E-3
PROBLEMS:
1. Find the [H+] if the pH is 12.4356. (3.66e-13M)
2. FIND THE pH IF THE [H+] IS .0000411. (4.3862)
ACID-BASE WORKSHEET II NAME______
I. WRITE THE NEUTRALIZATION EQUATIONS FOR: (BE SURE TO BALANCE)
a. NITRIC ACID AND LITHIUM HYDROXIDE 1111
b. SULFURIC ACID AND CALCIUM HYDROXIDE 1121
c. ACETIC ACID AND AMMONIUM HYDROXIDE 1111
II. WHAT IS THE [H+], [OH-], AND THE pH OF A SOLUTION MADE BY
ADDING 1.2 L. OF WATER TO 300 ml. OF .5 M HCL SOLUTION.
[H+]=.1 M
[Cl-]=.1 M
[OH-]=1E-13
pH = 1
III. WHAT IS THE [H+], [OH-], AND THE pH OF A SOLUTION MADE BY
ADDING 5.8 L. OF WATER TO 800 ml. OF .085 M HCL SOLUTION.
[H+]=.0103 M
[Cl-]=.0103 M
[OH-]= 9.71e-13
pH = 1.9872
IV. IF 15 GRAMS OF BARIUM HYDROXIDE ARE DISSOLVED INTO WATER TO MAKE 176 L. OF SOLUTION, FIND THE [Ba+2], [OH-], [H+] AND THE pH OF THE SOLN.
[Ba+2]=.000498
[OH-]=.000999 [H+]=1e-11
pH=11.0000
Acid-Base Worksheet III Name______Per_____
I. IF THE pH OF A NITRIC ACID SOLUTION IS 3.3040, FIND THE [H+], [NO3-]
AND THE [OH-].
[H+]=.000497
[NO3-]=.000497
[OH-]=2e-11
II. IF 200 ml OF .35 M HNO3 IS MIXED WITH 400 ml OF .2 M RbOH
CALCULATE WHICH CHEMICAL IS IN EXCESS, THE [H+], [OH-],
[Rb+], AND THE pH OF THE RESULTING SOLUTION.
EXCESS=.01 MOLES OH-
[OH-]=.0167 M
[H+]= 5.98e-13 M
pH= 12.2229
III. YOU ARE TITRATING 30 ml OF UNKNOWN ACID WITH .35 M KOH.
THE SOLUTION TURNS PINK AFTER 45.6 ml OF THE BASE IS ADDED.
CALCULATE THE [H+], [OH-] AND THE pH OF THE ACID SOLUTION.
pH=.2738 [H+]=.533 M [OH-]=1.87E-14
IV. IF 300 ml OF .45 M HNO3 IS MIXED WITH 400 ml OF .2 M Ca(OH)2
CALCULATE WHICH CHEMICAL IS IN EXCESS, THE [H+], [OH-],
[Ca+2], AND THE pH OF THE RESULTING SOLUTION.
EXCESS=.025 MOLES OH-
[OH-]=.0357 M
[H+]=2.8E-13 M
pH= 12.5527
V. YOU ARE TITRATING 25 ml OF UNKNOWN ACID WITH 1.5 M KOH.
THE SOLUTION TURNS PINK AFTER 8.2 ml OF THE BASE IS ADDED.
CALCULATE THE [H+] AND THE pH OF THE ACID SOLUTION.
[H+]=.492M pH=.3080
Acid-Base Worksheet IV Name ______Per ______
I. (Honors Only) 7.5 GRAMS OF THE ACID HA IS MIXED WITH WATER TO 2.5 LITERS.
HA DISSOCIATES INTO H+ IONS AND A- IONS. THE MOLAR MASS OF HA
IS 100 GRAMS/MOLE. FIND THE [H+], [A-], [HA], [OH-], & pH.
Ka=9.5E-9 (weak acid dissociation)
[HA]=2.99E-2 [H+]=1.69E-5 [OH-]=5.92E-10
II. WRITE THE NEUTRALIZATION EQUATIONS FOR: (BE SURE TO BALANCE)
a. PHOSPHORIC ACID AND LITHIUM HYDROXIDE 1331
b. SULFURIC ACID AND ALUMINUM HYDROXIDE 3261
c. ACETIC ACID AND BARIUM HYDROXIDE 2121
III. HOW MANY HYDROGEN IONS ARE IN 50 ml OF .025 M HBr SOLUTION?
7.53e20 ions
IV. WHAT IS THE MOLARITY OF HYDROGEN IONS IN AN ACID SOLUTION
WHOSE pH IS LISTED AS 1.3342? [H+]=4.63E-2M
V. IF 150 ml OF .035 M HI ARE MIXED WITH 150 ml OF .015 M Ca(OH)2,
FIND ALL ION CONCENTRATIONS IN RESULTING SOLUTION. ASSUME
BOTH CHEMICALS ARE STRONG ELECTROLYTES.
[I-]=1.75E-2
[Ca++]=7.5E-3
[H+]=2.5E-3
[OH-]=4E-12
pH=2.6021
VI. a.) WHAT IS THE MOLARITY OF AN UNKNOWN ACID IF 35 ml OF IT ARE
TITRATED TO EQUIVILENCE BY 17.3 ml OF .15 M NaOH.
7.41E-2M
b.) HOW MANY GRAMS OF SOLID SODIUM HYDROXIDE WILL BE NEEDED TO
MAKE 100 ml OF A BASE SOLUTION THAT IS AS STRONG AS THE
UNKNOWN ACID? 2.97E-1g
VII. (Honors only)A 25 GRAM SAMPLE OF BENZOIC ACID (C6H5COOH)IS DISSOLVED IN WATER
TO 1200 ml. FIND THE [SPECIES] AT EQUILIBRIUM. Ka=6.5E-6
FIND THE pH ALSO. X=1.05E-3M
[C6H5COOH] = 1.699e-3M