Addition of 0.070 mol NaOHCH
3NH
3Cl + NaOH
NaCl + CH
3NH
2 + H
2O
Adding NaOH would consume CH
3NH
3Cl, and therefore shift the reaction to the left.
[CH
3NH
2] = 0.80 mol + 0.070 mol = 0.87 mol
[CH
3NH
3Cl] = 1.00 mol - 0.070 mol = 0.93 mol
pH = 14-(-log(4.4x10
-4) + log(0.87 mol / 0.93 mol)) = 10.67
Addition of 0.11 mol HClCH
3NH
2 + HCl
CH
3NH
3+ + Cl
-Adding HCl would consume CH
3NH
2 to produce CH
3NH
3+, and therefore shift the reaction to the left.
[CH
3NH
2] = 0.80 mol - 0.11 mol = 0.69 mol
[CH
3NH
3+] = 1.00 mol + 0.11 mol = 1.11 mol
pH = -log(2.7x10
-11) + log(0.69 mol / 1.11 mol) = 10.36
Does this look correct?