pH and percent dissociation

## at 25 C, a 0.010mol/L ammonia solution is 4.3 percent ionized. Calculate the pH

# 2 Answers

OK - I like to focus on the reaction and what is happening - ammonia (NH3) is a weak base so it will only partially ionize as shown below

NH3(aq) + H2O(l) <----> NH_{4+ }(aq) + OH^{-}(aq)

So, if we start with 0.01 M NH3, what does it exactly mean when it is 4.3% ionized? This means that out of the original 0.01 M sample of NH3, only 4.3% is converted to NH4^{+}

Amount of NH3 converted to NH4^{+}: (4.3/100)(0.01) = 0.00043 M

So if we construct the following chart for the original reaction

NH3(aq) + H2O(l) <----> NH4+ (aq) + OH-(aq)

Initial molarity 0.01 mole/L 0 mole/L 0 mole/L

change in molarity - 0.00043 mole/L +0.00043 M 0.00043 mole/L

molarity at equilibrium 0.00957 M 0.00043 M 0.00043 M

Note: the change in the molarity for OH- is the same as the change in molarity for NH4^{+} because they have the same reaction coefficients

pH really depends on the H+ molarity

This is a useful equation: 1x10^{-14} = [H^{+}][OH^{-}] = [H+][0.00043 M]

[H+] = 1x10-14 / 0.00043 = 2.3x10^{-11}

pH = -log[H^{+}] = -log(2.3x10-11)

Hi Charles,

The reaction is NH_{3} + H_{2}O <==> NH_{4}^{+} + OH^{-}

To solve this, you use the Henderson-Hasselbalch equation,

pH = pK_{a} + log ([A]/[HA]) or pH = pK_{a} + log([NH_{3}]/[NH_{4}^{+}])

The pK_{a} at 25C for NH_{4}^{+} is 9.43 (you have to look this up).

We know that the solution is 4.3% ionized, so [NH_{4}^{+}] = 0.043*0.010 = 0.00043M

[NH_{3}] is 0.010M - 0.00043M = 0.00957M

Plugging these into the Henderson-Hasselbalch equation we get:

pH = pKa + log([NH_{3}]/[NH_{4}^{+}])

= 9.43 + log(0.00957/0.00043)

= 9.43 + log(22.3)

= 9.43 + 1.35

pH = 10.78

Another way to look at this is by simply using the ratios of NH_{3} and NH_{4}^{+}. Since you know in this particular case that the solution is 4.3% ionized, then NH_{4}^{+} will be 4.3% or 0.043 of the total. NH_{3}
then will be 1 - 0.043 = 0.957.

Plugging these into the Henderson-Hasselbalch equation we get:

pH = pK_{a} + log([NH_{3}]/[NH_{4}^{+}])

= 9.43 + log(0.957/0.043)

= 9.43 + log(22.3)

= 9.43 + 1.35

= 10.78

Same answer! The original concentration of the ammonia solution doesn't matter as long as you can calculate [A]/[HA].

Hope this helps.

Tom

## Comments

Another way you can look at this is by simply using the percentages. Since we know in this case that it is 4.3% ionized, [NH

_{4}^{+}] will be 4.3% or 0.043 of the total. Then [NH_{3}] is 1 - 0.043 = 0.957 of the total.Plugging these into the Henderson-Hasselbalch equation we get

pH = pKa + log([NH

_{4}^{+}]/[NH_{3}])= 9.43 + log(0.957/0.043)

= 9.43 + log(22.3)

= 9.43 + 1.35

pH = 10.78

Same answer! It didn't really matter the exact concentration of the ammonia solution, just the ratio of [NH

_{3}]/[NH_{4}^{+}]].Another way to look at this is by simply using the percent ionization. Since we know in this case that it is 4.3% ionized, then NH4

^{+}will be 4.3% or 0.043 of the total. NH_{3}is then 1 - 0.043 = 0.957 of the total.Plugging these into the Henderson-Hasselbalch equation we get

pH = pK

_{a}+ log ([NH_{3}]/[NH_{4}^{+}])= 9.43 + log (0.957/0.043)

= 9.43 + log(22.3)

= 9.43 + 1.35

pH = 10.78

Same answer! The concentration of the ammonia solution doesn't matter as long as you can find the ratio of [A]/[HA].

Absolutely not! The H-H equation is designed to be used in the buffer region of an acid/base buffer. This is within one pH unit of the pKa of the weak acid. It makes assumptions that are faulty once you get out of that zone. Sandra's technique is the correct one.

If you compare Tom's number (10.78) with Sandra's number (10.63), you may think they're close enough, but that's actually a pretty big difference. One thing about working with logs is that the numbers before the decimal are not significant. Thus, both 10.78 and 10.63 have two significant digits. They don't even match to within one significant digit. It's like saying that 78 and 63 are close enough to the same number.

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