Question

Consider the titration of a 26.0 −mL sample of 0.175 M CH3NH2 with 0.150 M HBr. Determine each of the following.

1. the pH at 4.0 mL of added acid

2.the pH at the equivalence point

3.the pH after adding 6.0 mL of acid beyond the equivalence point

Answer #1

1)

millimoles of CH3NH2 = 26 x 0.175 = 4.55

millimoles of HBr = 4 x 0.150 = 0.6

CH3NH2 + HBr ----------------> CH3NH3+Br-

4.55 0.6 0

3.95 0 0.6

pOH = pKb + log [CH3NH3+Br- / CH3NH2]

pOH = 3.30 + log (0.6 / 3.95)

pOH = 2.48

pH + pOH = 14

**pH = 11.52**

2)

at equilvalece point :

volume of HBr = 26 x 0.175 / 0.150 = 30.33 mL

here only salt present .

salt concentration = C = 26 x 0.175 / (26 + 30.33)

= 0.081 M

here

pH = 7 - 1/2 [pKb + log C] = 7 - 1/2 [3.30 + log 0.081]

**pH = 5.90**

3 )

volume of acid = 6 + 30.3 = 36.3

millimoles of acid = 36.3 x 0.150 = 5.445

millimoles of base = 4.55

acid millimoles remains after reaction = 5.445 -4.55 = 0.895

acid concetration = 0.895 / (36.5 +26)

= 0.0144 M

pH = -log (0.0144)

**pH = 1.84**

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