Consider the titration of a 27.0 mL sample of 0.170 M CH3NH2 with 0.155 M HBr. D

Question

Consider the titration of a 27.0 mL sample of 0.170 M CH3NH2 with 0.155 M HBr. Determine each of the following.

Part A

the initial pH

Express your answer using two decimal places.

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Part B

the volume of added acid required to reach the equivalence point

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Part C

the pH at 6.0 mL of added acid

Express your answer using two decimal places.

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Part D

the pH at one-half of the equivalence point

Express your answer using two decimal places.

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Part E

the pH at the equivalence point

Express your answer using two decimal places.

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Part F

the pH after adding 4.0 mL of acid beyond the equivalence point

Express your answer using two decimal places.

Consider the titration of a 27.0 mL sample of 0.170 M CH3NH2 with 0.155 M HBr. Determine each of the following.

Part A

the initial pH

Express your answer using two decimal places.

SubmitMy AnswersGive Up

Part B

the volume of added acid required to reach the equivalence point

SubmitMy AnswersGive Up

Part C

the pH at 6.0 mL of added acid

Express your answer using two decimal places.

SubmitMy AnswersGive Up

Part D

the pH at one-half of the equivalence point

Express your answer using two decimal places.

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Part E

the pH at the equivalence point

Express your answer using two decimal places.

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Part F

the pH after adding 4.0 mL of acid beyond the equivalence point

Express your answer using two decimal places.

Consider the titration of a 27.0 mL sample of 0.170 M CH3NH2 with 0.155 M HBr. Determine each of the following.

Part A

the initial pH

Express your answer using two decimal places.

pH =

SubmitMy AnswersGive Up

Part B

the volume of added acid required to reach the equivalence point

V =   mL  

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Part C

the pH at 6.0 mL of added acid

Express your answer using two decimal places.

pH =

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Part D

the pH at one-half of the equivalence point

Express your answer using two decimal places.

pH =

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Part E

the pH at the equivalence point

Express your answer using two decimal places.

pH =

SubmitMy AnswersGive Up

Part F

the pH after adding 4.0 mL of acid beyond the equivalence point

Express your answer using two decimal places.

pH =

Consider the titration of a 27.0 mL sample of 0.170 M CH3NH2 with 0.155 M HBr. Determine each of the following.

Part A

the initial pH

Express your answer using two decimal places.

pH =

SubmitMy AnswersGive Up

Part B

the volume of added acid required to reach the equivalence point

V =   mL  

SubmitMy AnswersGive Up

Part C

the pH at 6.0 mL of added acid

Express your answer using two decimal places.

pH =

SubmitMy AnswersGive Up

Part D

the pH at one-half of the equivalence point

Express your answer using two decimal places.

pH =

SubmitMy AnswersGive Up

Part E

the pH at the equivalence point

Express your answer using two decimal places.

pH =

SubmitMy AnswersGive Up

Part F

the pH after adding 4.0 mL of acid beyond the equivalence point

Express your answer using two decimal places.

pH =

Explanation / Answer

suppose base CH3NH2 = B

Kb of Base = 4.2 x 10^-4

pKb = -log Kb = -log (4.2 x 10^-4 ) = 3.38

(a) before addition of any HBr

B + H2O -------------------> BH+ + OH-

0.170                                 0          0 ---------------> initial

0.170-x                               x             x --------------> equilibrium

Kb = [BH+][OH-]/[B]

4.2 x 10^-4 = x^2 / 0.170-x

x^2 + 4.2 x 10^-4 x - 7.14 x 10^-5 = 0

x = 8.24 x 10^-3

x= [OH-] = 8.24 x 10^-3M

pOH = -log[OH-]

pOH = -log (8.24 x 10^-3)

pOH = 2.08

pH + pOH = 14

pH = 11.92

(b) volume added equivalence point

millimoles of B = 27 x 0.170 = 4.59

millimoles of acid = V x 0.155 = 0.155 V

at equivaelce point

0.155 V = 4.59

V = 29.6 mL

volume = 29.6

(c) addition 6 ml acid

millimoles of acid = 6 x 0.155 = 0.93

B      +    H+ -------------------> BH+

4.59      0.93                             0

3.66           0                             0.93

pOH = pKb + log [0.93 / 3.66]

pOH = 3.38 +log [0.93 / 3.66]

pOH = 2.785

pH = 11.22

part D )

half equivalence point

pKb = pOH

pOH = 3.38

pH   = 10.62

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