Consider the titration of a 270 mL sample of 0180 M CH3NH2 w

Consider the titration of a 27.0 mL sample of 0.180 M CH3NH2 with 0.155 M HBr. Find the following

1. initial pH

2. the volume of added acid required to reach the equivalence point

3. the pH at 4.0 mL of added acid

1) initially only CH3NH2 present

for such bases

pOH = 1/2 [pKb - log C]

pKb of CH3NH2 = 3.38

pOH = 1/2 [3.38 - log 0.180]

pOH = 2.06

pH = 14 - 2.06

pH = 11.94

2) millimoles of base = 27 x 0.180 = 4.86

4.86 millimoles HBr must be added

4.86 = V x 0.155

V = 31.35 mL

35.35 mL HBr required to reach equivalence point.

millimoles of HBr added = 4.0 x 0.155 = 0.62

4.86 - 0.62 = 4.24 millimoles CH3NH2 left

0.62 millimoles CH3NH3⁺ will form

total volume = 27 + 4 = 31.0 mL

[CH3NH2] = 4.24 / 31 = 0.137 M

[CH3NH3⁺] = 0.62 / 31 = 0.02 M

pOH = pKb + log [CH3NH3⁺] / [CH3NH2]

pOH = 3.38 + log [0.02] / [0.137]

pOH = 2.54

pH = 14 - 2.54

pH = 11.46