Acid- Base: pH/pOH Calculations

Strong Acids
If it is not a stong acid, it is a weak acid.
Know the strong acids (monoprotic and diprotice)
Strong acids ionize virtually 100%

HCl(aq)  + H2O => H3O+(aq)  +  Cl-(aq)

pH = -log[H3O+]; 10^-pH = [H3O+];


Question: Determine the pH of a 0.150M HCl solution
(since strong acid ionizes 100%, [HCl] = [H3O+]=0.150M

pH = -log(0.150M) = 0.824
Notice: 3 sgfigs in concentration 0.150M; 3 decimal places
 in pH value 0.824

Determine the concentration of a monoprotic acid solution if the pH
of the solution is 4.35
10^-4.35 =4.5x10^-5  M
Notice: 2 decimal places n pH value, 2 sigfigs in concentration answer

Only the decimal places of pH values are significant.


In a similar way that pH/[H3O+] connect, so can
pOH and [OH-]

pOH = -log[OH-] and 10^-pOH = [OH-]

Connecting pH and pOH?

@25C:   pH + pOH = 14.00

Question: Determine the pH of a 0.250M NaOH solution
Strong bases dissociate virtually 100%
NaOH is a strong base, dissociates 100%
NaOH(aq) => Na+(aq)  +   OH-(aq)
So, [NaOH] = [OH-] = 0.250M

When working with a base, one cannot determine pH directly from the given base concentration . Determine pOH, then determine pH

pOH = -log(0.250M) = 0.602
pH = 14.00-0.602 = 13.398

there is another way to carry out this calculation
- saving that for another time

A few basic jokes to end this post: What happened when the lion ate the comedian? Ans: It felt funny. Where do sheep go to get a haricut? Ans: To the Baa Baa shop. They are simple, basic jokes. Get it? This post dealt with bases, so it ended with some basic jokes!

llbm Review

-complete the llbm & solubility review handout
-check moodle for introduction to acid/base equilibrum handouts
-check moodle for upcoming quiz/test dates

Will a precipitate occur?

Will 35.0mL of 0.0100M NaCl form a precipitate when combined
with 65.0 mL of 0.0150M AgNO3?

• write the double displacement to identify the slightly soluble salt (the precipitate)
• silver chloride in the example
• determine the concentration of the ions of the slightly soluble salt using c1V1=c2V2
• Cl and Ag in this example
• for Cl: (0.0350L)(0.0100M)=c2(0.100L*);
• * 35mL + 65mL = 100mL = 0.100L
• Write the llbm equation for the slightly soluble salt then the Ksp expression
• Use the calculated concentrations, sub into Ksp to determine a "trial K"
• compare "trial K" to real K (from Ksp chart)
• if trial K less than real K = no precipitate (not enough of the ions present to cause a ppt to form)
• if trial K greater than real K = ppt (too great of an amount of the ions in the solution - excess ppts out)

Remember if you are not part of the solution, you are part of the precipitate.

Lab Follow-Up

The lab involved neutralizing 25.00mL of an unknown concentration of a calcium hydroxide solution with a determined volume (via the buret) of 0.100M hydrochloric acid.

The solution is already at equilibrium when the titrating begins.
With the titration data it is possible to:

• determine molar solubility of the calcium hydroxide in pure water
• determine Ksp value for calcium hydroxide
• determine molar solubility of calcium hydroxide in a calcium chloride solution

There was excess calcium hydroxide in the solution. the mass of calcium hydroxide is not relevant to the lab follow-up calculation. Furthermore, the calcium chloride was added to upset equilibrium. However, the mass of added calcium chloride is not relevant to the lab follow-up calculations.

The titration data is all that is needed to complete the calculations.


Next up - Acid-Base llbm. Two introductory sheets are on moodle. Notice the moodle calendar for upcoming quiz dates.


Talent alone won't make you a success. Neither will being in the right place at the right time, unless you are ready. The most important question is: 'Are your ready?' Johnny Carson Are you ready for acid-base solution equilibrium?