Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: cabaal on April 07, 2010, 08:07:39 PM
-
Could someone please check my work?
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. ::)
-
Can't you check it on your own?
-
I didn't know that tool existed. Thank you Borek.