Chemistry
Acid – Base Notes
• pH is a measure of a solution’s acidity or basicity.
• Acid: pH 0‒7
• Base pH 7‒14.
• Neutral pH = 7.0
• Stronger acid ---> lower pH.
• Stronger base ---> higher pH.
• Acids have an excess of HYDRONIUM ions, H3O+.
• Bases have an excess of HYDROXIDE ions, OH‒.
•Examples of strong acids:
•HCl – hydrochloric acid
•HNO3 – nitric acid
•H2SO4 – sulfuric acid
•Notice the formulas for these compounds all begin with H.
This is typical of acids, as the hydrogen ion H+ gives them their
acidic properties. Note: A hydrogen ion is actually a single
proton.
•Strong acids completely dissociate in water solution to become
100% ionized, e.g. HCl ---> H+ + Cl‒. Now this solution is composed of
nothing but positive and negative ions.
•A hydrogen ion in aqueous solution combines with a water molecule
to form a hydronium ion: H+ + H2O ---> H3O+
•The molar concentration of H3O+, written [H3O+], determines the
pH of an acid solution.
•pH = –log [H3O+]; This reads: pH equals the negative logarithm of
the hydronium ion concentration.
•Example:
Determine the pH of an acid solution whose [H3O+] = 2.0 x
10‒4.
•pH = ‒log (2.0 x 10‒4)
•On your TI calculator enter:
‒
LOG
pH = 3.70
2.0
2nd
EE
‒
4
)
enter
•Examples of strong bases:
•NaOH – sodium hydroxide
•KOH – potassium hydroxide
•Hydroxide ions are either directly contributed or are generated by
the compound in aqueous solution:
NaOH ---> Na+ + OH – (strong base)
• Like strong acids, strong bases completely ionize in aqueous
solution. In the example above, the solution contains 100% sodium
& hydroxide ions; none of the compoun remains.
• An example of a weak base generating an OH – ion:
NH3 + H2O ---> NH4+ + OH –
In this case the ammonia gained a hydrogen ion from water,
leaving a remaining hydroxide ion along with an ammonium ion.
• The molar concentration of hydroxide ion, written [OH –], determines
the pOH of a base solution.
• pOH = –log [OH –]; This reads: pOH equals the negative logarithm of
the hydroxide ion concentration.
• Example:
Determine the pOH of a base solution whose [OH –] = 3.5 x 10‒2.
• pOH = ‒log (3.5 x 10‒2)
• On your TI calculator enter:
‒
LOG
3.5
2nd
EE
‒
2
)
Enter
pOH = 1.46
IMPORTANT: pH + pOH = 14
• So the above solution has a pH of 14 – 1.46 = 12.54 (clearly a basic
solution.)
Examples:
1a. Determine the pH of an acid solution of [H3O+] = 1.00 x 10‒3 M
pH = ‒log [H3O+] = ‒log 1.00 x 10‒3
‒
LOG
1.00
2nd
EE
‒
3
)
enter
pH = 3.0
1b. What is the pOH of this solution?
pH + pOH = 14; pOH = 14 – pH = 14 – 3.0; pH = 11.0
2a. A base solution has a concentration of [OH‒] = 2.59 x 10‒5;
calculate the pOH.
pH = ‒log [OH‒] = ‒log 2.59 x 10‒5
‒
LOG
2.59
2nd
EE
‒
pOH = 4.59
2a. What is the pH of this solution?
pH = 14 – pOH = 14 – 4.59; pH = 9.41
5
)
enter
3. What is the [H3O+] and pH of a 4.21 M solution of HNO3?
a. Since HNO3 is a strong acid, it exists in solution as ions only:
HNO3 ---> H+ + NO3 –
So 4.21 M HNO3 ---> 4.21 M H+ + 4.21 M NO3 –
4.21 M H+ + H2O ---> 4.21 M H3O+; [H3O+] = 4.21 M
3b. pH = ‒log [H3O+] = ‒log 4.21
‒
LOG
4.21
)
enter
pH = ‒0.62
NOTE: If your calculated result is negative, round off to ZERO.
So, pH = 0.
3c. What is the pOH? pOH = 14 – pH; pOH = 14 – 0 = 14
4. What is the [H3O+] of a solution whose pH = 4.75?
Now we must do the inverse log calculation.
pH = ‒log [H3O+]
‒pH = log [H3O+]
[H3O+] = log –1 (‒pH)
= log –1 (‒4.75)
2nd
LOG
‒4.75
[H3O+] = 1.78 x 10 –5 M
Enter