Question

Calculate the pH for each of the following cases in the titration of 35.0 ml of 0.200 M NaOH(aq), with 0.200 M HBr(aq).

Answer 1

a) pH = 14-pOH pOH = -log(0.2) = 0.70 pH = 14 - 0.7 = 13.3

b) moles of NaOH = 0.2x0.035 = 0.007 moles   moles of HBr = 0.2x0.0135 = 0.0027 moles

moles of NaOH in equilibrium: 0.007-0.0027 = 0.0043 moles

[OH] = 0.0043 / (0.035+0.0135) = 0.089 M pH = 14 - log(0.089) = 12.95

c) moles of NaOH = 0.2x0.035 = 0.007 moles   moles of HBr = 0.2x0.0255 = 0.0051 moles

moles of NaOH in equilibrium: 0.007-0.0051 = 0.0019 moles

[OH] = 0.0019 / (0.035+0.0255) = 0.031 M pH = 14 - log(0.031) = 12.50

d) moles of NaOH = 0.2x0.035 = 0.007 moles   moles of HBr = 0.2x0.035 = 0.007 moles

moles of NaOH in equilibrium: 0.007-0.007 = 0   pH = 7

e) moles of NaOH = 0.2x0.035 = 0.007 moles   moles of HBr = 0.2x0.0455 = 0.0091 moles

moles of NaOH in equilibrium: 0.007-0.0091 = -0.0021 moles

[HBr] = 0.0021 / (0.0035+0.0455) = 0.026 M pH = - log(0.026) = 1.58

f) moles of NaOH = 0.2x0.035 = 0.007 moles   moles of HBr = 0.2x0.05 = 0.01 moles

moles of NaOH in equilibrium: 0.007-0.01 = -0.003 moles

[HBr] = 0.003 / (0.0035+0.05) = 0.035 M pH = - log(0.035) = 1.45