Consider the titration of a 26.0 -- mL sample of 0.175 M CH3NH2 with 0.155 M HBr

Question

Consider the titration of a 26.0 -- mL sample of 0.175 M CH3NH2 with 0.155 M HBr.

Initial pH= 11.94

volume of added acid required to reach the equivalence point: 29.4 mL

the pH at 4.0 mL of added acid: 11.45

the pH at one-half of the equivalence point: 10.64

Part E the pH at the equivalence point Express your answer using two decimal places. pH Submit My Answers Give Up Part F the pH after adding 6.0 mL of acid beyond the equivalence point Express your answer using two decimal places. pH Submit My Answers Give Up

Explanation / Answer

part E)

total volume at equivalence point = 26 + 29.4 = 55.4 mL

[CH3NH3+] = (26 x 0.175) / 55.4 = 0.082 M

pOH = 1/2 [pKw + pKb + logC]

pKb = 3.38 standard value

pOH = 1/2 [14 + 3.38 + log 0.082]

pOH = 8.14

pH = 14 - 8.14

pH = 5.86

partF) millimoles of acid added = (29.4 + 6.0) x 0.155 = 5.487

millimoles of base present = 26 x 0.175 = 4.55

5.487 - 4.55 = 0.937 millimoles acid left

[HBr] = 0.937 / 61.4 = 0.0152 M

pH = - log [H+]

pH = - log [0.0152]

pH = 1.82

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