ph calculations

pH Calculations

pH and pOH Solutions

1) What is the pH of the solution with a hydronium concentration [H3O+] 1.47 x 10-4?

What is the pOH of this solution?

pH = -log (1.47 x 10-4) = 3.83

pOH = 14 - pH = 14 - 3.83 = 10.17

2) What is the pOH of the solution with a hydroxyl concentration [OH-] 2.98 x 10-2?

What is the hydronium concentration [H3O+] of this solution?

pOH = -log (2.98 x 10-2) = 1.53

hydronium concentration [H3O+] = 1 x 10 -14 = 3.37 x 10-13

2.98 x 10-2

OR pH = 14- pOH = 14 - 1.53 = 12.47

hydronium concentration [H3O+] = antilog -12.47 = 3.38 x 10-13

3) What is the hydronium concentration [H3O+] of a solution with a pH of 7.84?

What is the hydroxyl concentration [OH-] of this solution?


hydroxyl concentration [OH-] = 1 x 10 -14 = 6.90 x 10-7

1.45 x 10-8

OR pOH = 14-pH = 14 - 7.84 = 6.16

hydroxyl concentration [OH-] = antilog -6.16 = 6.9 x 10-7

4) What is the hydroxyl concentration [OH-] of a solution with a pH of 3.76?

pOH = 14 - pH = 14 - 3.76 = 10.24

hydroxyl concentration [OH-] = antilog -10.24 = 5.75 x 10-11

5) What is the hydronium concentration [H3O+] of a solution with a pOH of 2.47?

pH = 14 - pOH = 14 - 2.47 = 11.53


Strong Acid and Base: are the ones that dissociate completely

1. The 100% dissociation idea begins to break down as solutions become more concentrated. Usually if the acid is 100% dissociated in solutions of 1.0-molar or less, it is called strong.

2. Sulfuric acid is considered strong only in its first dissociation step. (Here's a possible problem for the future: you are given a problem which treats the sulfuric acid as 100% ionzed in BOTH dissociation steps. What to do? Do the problem in the manner desired. Your choice as to discussing it with your instructor.)

0.1 M of strong Acid will give 0.1 M of [H3O+] ions

Weak Acid and Base:

Weak electrolytes will dissociate in solution, but they do so less than 100%. The usual examples that student study will dissociate only 1 to 5%.

There is a constant that controls the dissociation ka

Ex: 0.1m of CH3COOH will dissociate with ka=1x10-2 at 25C

CH3COOH +H2O H3O+ + CH3COO-

Initially

0.1 + 0 0 + 0

After dissociation

0.1-x x + x

ka=¿¿

x2+0.01x−0.01=0

x=−0.01+√0.0001+0.042

=0.095

pH=−log¿¿

Problem #1: A weak acid has a pKa of 4.994 and the solution pH is 4.523. What percentage of the acid is dissociated?

A comment before discussing the solution: note that the pKa is given, rather than the Ka. The first thing we will need to do is convert the pKa to the Ka. Then, the two values we need to obtain to solve the problem given just above are [H+], which is pretty easy and [HA], which is only a tiny bit more involved.

Solution:

1) Convert pKa to Ka:

Ka = 10¯pKa = 10¯4.994 = 1.0139 x 10¯5

Often the identity of the weak acid is not specified. That is because, with few exceptions, all weak acids behave in the same way and so the same techniques can be used no matter what acid is used in the problem. In cases where no acid is identified, you can use a generic weak acid, signified by the formula HA. Here is the dissociation equation for HA:

H+ + A¯ <===> HA

Here is the equilibrium expression for that dissociation:

Ka = ([H+] [A¯]) / [HA]

2) The pH will give [H+] (and the [A¯]):

[H+] = 10¯pH = 10¯4.523 = 3.00 x 10¯5 M

Because of the 1:1 molar ratio in the above equation, we know that [A¯] = [H+] = 3.00 x 10¯5 M.

3) This means that the only value left is [HA], so we will use the equilibrium expression to calculate [HA].

1.0139 x 10¯5 = [(3.00 x 10¯5) (3.00 x 10¯5)] / x

x = 8.88 x 10¯5 M

4) Percent dissociation for an acid is [H+] / [HA] and then times 100.

3.00 x 10¯5 / 8.88 x 10¯5 = 33.8%

Problem #2: A solution of acetic acid (Ka = 1.77 x 10¯5) has a pH of 2.876. What is the percent dissociation?

Solution:

1) Calculate the [H+] from the pH:

[H+] = 10¯pH = 10¯2.876 = 1.33 x 10¯3 M

2) From the 1:1 stoichiometry of the chemical equation, we know that the acetate ion concentration, [Ac¯] equals the [H+]. Therefore,

[Ac¯] = 1.33 x 10¯3 M

3) We need to determine [HAc], the acetic acid concentration. We use the Ka expression to determine this value:

1.77 x 10¯5 = [(1.33 x 10¯3) (1.33 x 10¯3)] / x

x = 0.09993 M = 0.100 M

4) Percent dissociation:

(1.33 x 10¯3 / 0.100) times 100 = 1.33%

Comment: the first example is somewhat artifical, in that the percent dissocation is quite high. The second example is more in line with what teachers usually ask. The usual percent dissociation answer is between 1 and 5 per cent. However, the 33.8% answer, while not commonly found in introductory chemistry classes, is possible.

Problem #3: A generic weak acid (formula = HA) has a pKa of 4.401. If the solution pH is 3.495, what percentage of the acid is undissociated?

Solution:


Ka = 10¯pKa = 10¯4.401 = 3.97 x 10¯5

2) The pH gives [H+] (and the [A¯]):

[H+] = 10¯pH = 10¯3.495 = 3.20 x 10¯4 M

3) Determine the concentration of the weak acid:

Ka = ([H+] [A¯]) / [HA]

3.97 x 10¯5 = [(3.20 x 10¯4) (3.20 x 10¯4)] / x

x = 0.00258 M

4) Determine percent dissociation:

3.20 x 10¯4 / 0.00258 = 12.4%

5) Determine percent undissociated:

100 - 12.4 = 87.6%

Comment: the calculation technique discussed above determines the percent dissociation. Notice that the above problem asks for the percent undissociated.

Be aware! The problem above goes one step beyond what is normally taught. This might show up as a test question.

Problem #4: A weak acid has a pKa of 4.289. If the solution pH is 3.202, what percentage of the acid is dissociated?

Solution:


Ka = 10¯pKa = 10¯4.289 = 5.14 x 10¯5

2) The pH gives [H+] (and the [A¯]):

[H+] = 10¯pH = 10¯3.202 = 6.28 x 10¯4 M

3) Determine the concentration of the weak acid:

Ka = ([H+] [A¯]) / [HA]

5.14 x 10¯5 = [(6.28 x 10¯4) (6.28 x 10¯4)] / x

x = 0.00767284 M (kept a few guard digits on this one

4) Determine percent dissociation:

6.28 x 10¯4 / 0.00767284 = 8.2%

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