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13 Dec 2019

Consider the titration of a 25.0 −mL sample of 0.110 M HC7H5O2 with 0.120 M KOH. Determine each of the following.

The Ka of HC7H5O2 is 6.5 × 10−5.

Part A: the pH at 5.00 mL of added base pH= ?

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Elin Hessel
Elin HesselLv2
17 Dec 2019

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Related questions

Calculate the pH of a solution formed by mixing 150.0 mL of 0.10 M HC7H5O2 with 100.0 mL of 0.30 M NaC7H5O2. The Ka for HC7H5O2 is 6.5 x 10-5.

4.31
4.49
10.51
4.19
9.69

(1) A 40.0 mL sample of 0.134 M HNO2 is titrated with 0.262 M KOH. (Ka for HNO2 is 4.57×10−4.) Determine the pH at the equivalence point for the titration of HNO2 and KOH

(2) Consider the titration of a 25.0-mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH. (The value of Ka for HC2H3O2 is 1.8×10−5.) Determine the pH after adding 4.0 mL of base beyond the equivalence point.

(3) Consider the titration of a 25.0-mL sample of 0.175 M CH3NH2 with 0.145 M HBr. (The value of Kb for CH3NH2 is 4.4×10−4.) (a) Determine the pH at one-half of the equivalence point. (b) Determine the pH after adding 5.0 mL of acid beyond the equivalence point.

Consider the titration of a 25.0-mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH. (The value of Ka for HC2H3O2 is 1.8×10−5.)

Found:

initial Ph =2.85

Determine the volume of added base required to reach the equivalence point. = 21.2 ml

Find

part A

Determine the pH at 4.0 mL of added base.

Part B

Determine the pH at one-half of the equivalence point.

Part C

Determine the pH at the equivalence point.

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