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) NH
3 (.1 mol/2 L) NH
4+[NH
4+] = 0.05 M
NH
4+ + H
2O
NH
3 + H
3O
+0.05 - - -
0.05-x - +x +x
x
2/(0.05-x) = K
A = 5.6*10
-10 ==> [H
3O
+] = 5.29*10
-6 M
pH = -log([H
3O
+]) = 5.28
B.
(.1 mol/1 L) CH
3COOH + (.1 mol/1 L) NaOH
(.1 mol / 2 L) CH
3COO
- + H
2O
[CH
3COO
-] = 0.05 M
CH
3COO
- + H
2O
CH
3COOH + OH
-0.05 - - -
0.05-x - +x +x
x
2/(0.05-x) = K
b = 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.