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General : Repost: Acid-Base Reactions
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 Message 1 of 2 in Discussion 
From: MSN Nickname·Steve·  (Original Message)Sent: 11/7/2008 6:28 PM
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Recommend Delete    Message 5 of 6 in Discussion 
From: MSN NicknameAlbert1145 Sent: 11/7/2008 12:04 AM
I came across two acid and base equations last nigh, pretty similar but I can't do them. I think I have some problems with writing out chemical equation and the ICE table, could you help me with these two also?

10.3 mL 0.200 M nitric acid solution is added to 50.00 mL 0.100 M sodium hydroxide solution. What is the pH of the solution?

24.2 mL 0.200 M nitric acid solution is added to 50.00 mL 0.200 M ammonia solution. What is the pH of the solution? (Kb for ammonia 1.80 x 10-5)


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 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·Sent: 11/7/2008 6:55 PM
First, it's always a good idea to write the balanced chemical reactions:
 
1)  HNO3 (aq)  +  NaOH (aq)   ––�?gt;  NaNO3 (aq)  +  H2O (l)
     Strong Acid    Strong Base          Neutral Salt
 
2)  HNO3 (aq)  +  NH3 (aq)   ––�?gt;   NH4NO3 (aq)
     Strong Acid    Weak Base         NH4+ = Weak Acid (NO3�?/SUP> = neutral)
 
 
For both questions, first calculate the moles of each reactant.  Which is the limiting reactant?  How many moles of the excess reactant will remain?
 
For Question 1, that's all you need.  Either HNO3 or NaOH will remain in excess.  The moles divided by the total volume gives the molarity of H+ if HNO3 was in excess, or OH�?/SUP> if NaOH was in excess.  From this, you can calculate the pH.
 
For Question 2, you also need to calculate how many moles of NH4NO3 are formed.  It will turn out in this problem that HNO3 is the limiting reactant, so after reaction there will be excess NH3 remaining, plus the product NH4NO3.  NH4+ is the important part of the product, because it is the conjugate acid of NH3 and a weak acid also.  NO3�?/SUP> is the conjugate base of a strong acid, and as such, it is such a weak base that it is neutral.
 
Thus you will have in this solution a weak base (NH3) plus its conjugate acid (NH4+).  That is a buffer solution.  A "shortcut" for calculating the pH of buffer solutions is the Henderson-Hasselbalch equation:
 
pH  =  pKa  +  log{[weak base] / [conjugate acid]}
 
or, pH  =  –log(Ka)  +  log{[NH3] / [NH4+]}
                        ^
Note that you have to convert Kb to Ka.
 
You can also take the "ICE" approach, a little longer, but it gives exactly the same result.  But first, go ahead and work Question 1, to be sure to get the "bugs" worked out, and then tackle Question 2. 
 
Steve