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Acid-Base Equilibria & Neutralization

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Acid-Base Equilibria & Neutralization

Sorry, I didn`t find chemistry section on this forum. I have a question regarding neutralization of HCl with NH4OH (or NH3) weak base.

In lab we neutralized 50 mL of 2.0 M NH3 (weak base) with 50 mL of 2.0 M HCl (strong acid). I can`t find how many moles of water was produced from this reaction. 0.1 mol of HCl will be used if it is mixed with a strong base (50 mL 2.0 M NaOH or 50 mL 2.0 M KOH for example) , but ammonia is a weak base, so it won`t completely dissociate.

Do we need to use the base dissociation constant called (Kb) to find out "n" moles of water produced?
samoyan
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For this pair, 1 or none depending on how you want to write the equation. Kb or Ka won't enter into it since you can assume that HCl will completely dissociate and that NH4OH will completely neutralize the excess protons (hydronium ions-if you prefer).
blcr11
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I should be more explicit. Depending on how you write the equation, you will generate 1 mole of H2O per mole of NH4OH neutralized. If you write it as the sum of "half-reactions" with water, the net reaction consumes as many waters as it generates--which is not surprising as this is an aqueous solution and the concentration of water is essentially fixed.
blcr11
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Thank you for the help.

According to first reactions of neutralization, the heat given off by the reaction was calculated:

Experimental results:

50 mL of 2.0 HCl , 50 mL of NaOH and 50 mL of 2.0 NaOH was used in these reactions.

HCl + NaOH --> NaCl + H2O (Q = - 4660.11 J)

HCl + KOH --> KCl + H2O (Q = - 5645.57 J)

We know that ratio is 1:1 and that both reactants dissociate completely in these reactions, so 0.1 moles of water is produced in these reactions. Knowing this, we can calculate ΔH.

For the first reaction:

ΔH = Q/n = - 4660.11 J / 0.1 mol = - 46601.1 J/mol = - 46.6 kJ/mol

For the second reaction:

ΔH = Q/n = - 5645.57 J / 0.1 mol = - 56455.7 J/mol = - 56.46 kJ/mol

Now, a third reaction:

A weak base NH4OH and HCl were used.
Calculated Q for this reaction = - 3883.19 J
As we can see, the Q for this reaction is lower than for the reactions of strong base + strong acid. So, we can assume that not all of 50 mL 2.0 M NH4OH was used to neutralize 50 mL 2.0 M HCl.

The key to solving this problem is to find out number of moles produced in the reaction. If we assume that all NH4OH was used in this reaction, we`ll get 0.1 moles of water produced. But why then the Q is so different between these reactions?
samoyan
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heat release = bonds broken - bonds formed
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Enjoying one moment at a time;
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mith
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Sorry. Wrong model. This is a calorimetric determination of heats of neutralization, which is something other than a simple end-point titration. The end-points are the same for strong or weak acids and bases, but the heats evolved are not necessarily the same. Here there can be an influence from the incomplete ionization of ammonium hydroxide and the formation of some small amount of solvated, neutral NH3. At least transiently, some of the protons from HCl have to protonate NH3 back to the ammonium ion and that will change the heats evolved compared to a reaction with NaOH or KOH. You should still completely neutralize the NH4OH, you just won't get as much heat out of the reaction.
blcr11
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blcr11, thank you very much for the explanation. I got it now.
samoyan
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