1) Consider the titration of a 36.0 mL sample of 0.180 M HBr with 0.210 M KOH. D

Question

1) Consider the titration of a 36.0 mL sample of 0.180 M HBr with 0.210 M KOH. Determine each of the following:

Part B

the volume of added base required to reach the equivalence point

Part C

the pH at 10.9 mL of added base

(initial pH is 0.745)

2) Consider the titration of a 23.4 mL sample of 0.115 M RbOH with 0.110 M HCl. Determine each quantity:

Part A

the initial pH

Part B

the volume of added acid required to reach the equivalence point

Part C

the pH at 4.0 mL of added acid

Part D

the pH at the equivalence point

Part E

the pH after adding 5.6 mL of acid beyond the equivalence point

Explanation / Answer

Part B

No of Moles in HBr= .036 X 0.18= 0.00648

Volume of added base required to reach equivalence point = 0.00648 / 0.210 = 0.03086 L= 30.86 mL

Part C

In 10.9 mL added base

No. of Moles in KOH= 0.0109 X 0.210= 0.002289

Moles of HBr which are excess= 0.00648-.002289= 0.004191

Total volume= 36 mL + 10.9 mL =46.9 mL = 0.0469 L

Concentration of H+ = 0.004191/0.0469 = 0.0894 M

pH at 10.9 mL of added base= -log 0.0894= 1.0487 = 1.05

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