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General : pH of solutions
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Message 1 of 4 in Discussion

From:

goldie647 (Original Message)

Sent: 3/14/2008 10:59 PM

Given 20ml, o.o5M acetic acid (Ka=1.8x10^5) titrated with 0.05M NaOH. Calculate the pH of the following solutions:

mixture after adding 25ml, 0.05M NaOH?

Can you use the Henderson Hasselbalch equation for this?

My professor used the following

[OH-] > [H+] -> diff [OH-] = Mb - Ma = ?M

pOH = -log(OH-) = ?

pH = 14 - pOH = ?

But I do not understand this theory.

Thanks

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Message 2 of 4 in Discussion

From:

·Steve·

Sent: 3/15/2008 12:07 AM

We should follow the usual steps, first calculating the moles of each reactant and determining which is the limiting reactant. Then we should be able to tell how to calculate the pH.

moles of acetic acid = 0.0010 mol
moles of NaOH = 0.00125 mol

The reaction is

HC₂H₃O₂ (aq) + NaOH (aq) ––�?gt; NaC₂H₃O₂ (aq) + H₂O (l)

Initially: 0.0010 mol 0.00125 mol 0 (solvent)
limiting excess

After Rxn: 0 0.00025 mol 0.0010 mol (solvent)

So what we have in the end is 0.00025 mol of NaOH and 0.0010 mol of NaC₂H₃O₂ in 45.0 mL of solution.

The NaOH, being a strong base, will primarily determine the OH^{�?/SUP> concentration. The hydrolysis reaction of NaC₂H₃O₂ will contribute a negligible amount of OH^{�?/SUP> by comparison, So, we can ignore any OH�?that results from the sodium acetate.}}

Your teacher is dividing the moles of acetic acid and NaOH by the total volume at the outset, whereas I have not divided by the total volume yet. I usually use moles when doing the stoichiometry calculations, and convert to molarity at the end, but either way is OK, you get the same result. The formula, "diff [OH^{�?/SUP>] = M_b �?/FONT> M_a = ? M" is the net result. Notice in the reaction, the moles of NaOH that remain as excess is the difference between the moles of NaOH and the moles of acetic acid. If we divide each of the moles by the total volume, 0.0450 L, we will have the initial molarities of the NaOH and acetic acid, after they have been mixed together but before they have reacted. The difference is the molarity of the excess NaOH that remains, which is 0.0056 M.}

So, the pOH is –log(0.0056 M) = 2.25.

By the way, if you are curious about how much OH^{�?/SUP> will be contributed by the sodium acetate, we can calculate that by plugging into the K_b expression for the acetate ion. The molarity of the acetate ion is 0.0010 mol / 0.045 L = 0.022 M.}

C₂H₃O₂^{�?/SUP> (aq) + H₂O (l) HC₂H₃O₂ (aq) + OH^{�?/SUP> (aq)}}

Initially: 0.022 M �?nbsp; 0 0.0056 M

At Equilib: 0.022 �?x �?nbsp; x 0.0056 + x

Kb = [HC2H3O2] [OH^�?/SUP>]

[C2H3O2^�?/SUP>]

5.6 X 10^�?0 = (x)(0.0056 + x) �? (x)(0.0056)

0.022 �?x 0.022

x = 2.2 X 10^�? M

This is indeed a negligible amount of hydroxide ion compared to 0.0056 M.

Steve

Message 3 of 4 in Discussion

From:

goldie647

Sent: 3/16/2008 8:17 PM

How do you know when you can use my professors way?

[OH-] > [H+] -> diff [OH-] = Mb - Ma = ?M

What pops out in the question that should make me consider this method?

Thanks

Message 4 of 4 in Discussion

From:

·Steve·

Sent: 3/17/2008 1:22 AM

Typically, if there is a significant excess of strong acid or strong base in the solution, then no matter what the other substances present are, the pH is primarily determined just by the concentration of strong acid or strong base.

In this problem, nothing popped out at me until I looked at the moles of acid and base initially present. Then I realized that excess NaOH was added, and there would therefore be some NaOH remaining in the solution after the reaction with the acetic acid.

Here are the other typical situations for which you may have to calculate the pH:

1) Solution of a strong monoprotic acid like HCl or HNO₃. [H⁺] = [strong acid] since dissociation of H⁺ is 100%.

2) Solution of a strong base like NaOH. [OH^{�?/SUP>] = [strong base] since dissociation of OH^{�?/SUP> is 100%.}}

3) Solution of a weak acid, "HA" at initial concentration [HA]_i. [H⁺] = (K_a[HA]_i)^½. This is true if the percent ionization is less than about 5% of the initial weak acid concentration. If the percent ionization is more than this, it is better to use the quadratic formula to calculate "x" (which is the H⁺ concentration).

4) Solution of a weak base "B" at initial concentration [B]_i. [OH^{�?/SUP>] = (K_b[B]_i)^½. This is true if the percent of the weak base that has undergone the weak base hydrolysis reaction is less than about 5% of the initial concentration of the weak base. If the percent hydrolysis is more, the quadratic formula should be used to calculate "x" (which is the OH^{�?/SUP> concentration).}}

5) Solution of a mixture of a weak acid and its conjugate base, or, a solution of a weak base and its conjugate acid. These are buffer solutions for which you can use the Henderson-Hasselbalch equation to calculate the pH.

Steve

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