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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: cabaal on April 07, 2010, 08:07:39 PM

Title: Calcuating pH equivalence point for titrations.
Post by: cabaal on April 07, 2010, 08:07:39 PM
Could someone please check my work?

Quote
Calculate the pH at the equivalence point for these titrations: (a) .10 M HCl versus 0.10 M NH3, (b) 0.10 M CH3COOH versus 0.10 M NaOH.

I'm assuming a 1 liter solution of each.

A.
(.1 mol/1 L) HCl + (.1 mol/1 L) NH3 :rarrow: (.1 mol/2 L) NH4+
[NH4+] = 0.05 M
NH4+ + H2O  ::equil:: NH3 + H3O+
0.05     -               -     -
0.05-x  -               +x   +x
x2/(0.05-x) = KA = 5.6*10-10 ==> [H3O+] = 5.29*10-6 M
pH = -log([H3O+]) = 5.28

B.
(.1 mol/1 L) CH3COOH + (.1 mol/1 L) NaOH  :rarrow: (.1 mol / 2 L) CH3COO- + H2O
[CH3COO-] = 0.05 M
CH3COO- + H2O  ::equil:: CH3COOH + OH-
0.05          -             -              -
0.05-x       -             +x            +x
x2/(0.05-x) = Kb = 5.6*10-10 ==> [OH-] = 5.29*10-6 M
pH = 14+log([OH-]) = 8.72


Yes, I'm aware there's a shortcut...but I'll probably end up forgetting it or messing it up on the day of the exam.  ::)
Title: Re: Calcuating pH equivalence point for titrations.
Post by: Borek on April 08, 2010, 03:23:14 AM
Can't you check it on your own?
Title: Re: Calcuating pH equivalence point for titrations.
Post by: cabaal on April 08, 2010, 09:31:48 AM
I didn't know that tool existed. Thank you Borek.