Question

When 100ml of 0.01 M ethyl acetate react with 100ml of 0.01 M NaOH the unit of the rate constant is

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Answer 1

.01=0.4

.01=0.4Clearly M

.01=0.4Clearly M 1

.01=0.4Clearly M 1

.01=0.4Clearly M 1 V

.01=0.4Clearly M 1 V 1

.01=0.4Clearly M 1 V 1

.01=0.4Clearly M 1 V 1 >M

.01=0.4Clearly M 1 V 1 >M 2

.01=0.4Clearly M 1 V 1 >M 2

.01=0.4Clearly M 1 V 1 >M 2 V

.01=0.4Clearly M 1 V 1 >M 2 V 2

.01=0.4Clearly M 1 V 1 >M 2 V 2

.01=0.4Clearly M 1 V 1 >M 2 V 2

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M=

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V 2

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V 2

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V 2

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V 2

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V 2 =

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V 2 = 140

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V 2 = 1401−0.4

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V 2 = 1401−0.4

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V 2 = 1401−0.4 =0.0042

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V 2 = 1401−0.4 =0.0042pH=−log(0.0042)

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V 2 = 1401−0.4 =0.0042pH=−log(0.0042) =2.37

.01=0.4Clearly M 1 V 1 >M 2 V 2 ∴ Resulting solution will be acidic.Molarity of resulting solution M= V 1 +V 2 M 1 V 1 −M 2 V 2 = 1401−0.4 =0.0042pH=−log(0.0042) =2.37∴ pH of solution =2.37

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Answer 2

Here is your answer:-

The correct answer for your question is 2.37

Refer to the attachment for details:-

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