D oscerfolio for C https://organicccstonybrookedu/che133-16/oscerlolio/chguzman

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D oscerfolio for C https://organicccstonybrookedu/che133-16/oscerlolio/chguzman php 18.4052 nitial weight of KHP vial (gl 154464 Final weight of KHP vial (g) Volume of KHP solution (mL) What is the concentration of the KHP solution? Note: The answers to almost al the subsequent Question 1 questions depend on the concentration of KHP.J A solution of sodium hydroxide was standardized by titrating known volumes of the above KHP solution. The resulting data are summarized in the table below: RUN RUN 2 Initial KHP buret reading 2,43 4.04 Final KHP buret reading 33.35 36.84 Initial NaoH buret reading 385 266 Final NaOH buret reading 22.53 22.38 What was the concentration of the NaOH in Run 1 of the standardization? Question 2 Enter Your Answer What was the concentration of the NaOH in Run 2 of the s andardization? Question 3 What was the average concentration of the NaOH? Note: The accuracy of the average concentration Question 4 of NaOH is critical in the answers to most of the subsequent questions.] Enter Your Answer: .0768 Incorrect What was the average deviation of the concentration of the NaOH? Question What was the percent deviation of the concentration of the NaOH? Question 6 2.7 The standardized NaOH solution was used to titrate vinegar samples. The vinegar was delivered from a 5.00 mL transfer pipet and diluted with water. The results of three titrations are given in the table below a O Ask me anything 14 of 15 Answers Have been Submitted 1 One submission Write PDF file. Open Print Friendly 113 PM

Explanation / Answer

Question 1:

[KHP] = Moles / V in L

Moles = Mass /MW = 18.4052 - 15.4464 / 204.22 = 0.014488

[KHP] = 0.014488 / 0.25 = 0.058

Question 2:

KHP + NaOH -----> H2O + KNaP

Moles of KHP = Moles of NaOH

M1V1 = M2V2

0.058 * (33.35-2.43) = M2*[22.53 - 3.85]

M2 = 0.0965

[NaOH] = 0.0965 M

Question 3

0.058*[36.84-4.04] = M2*[22.38-2.66]

M2 = 0.09647

[NaOH] = 0.09647 M

Question 4 :

Average [NaOH] = 0.09647 + 0.0965 / 2 = 0.096485

Question 5 :

Average deviation =[ 0.0965 - 0.096485] + [0.09647 - 0.096485] / 2 = 0

Question

Percent deviation= 0

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