In a titration, the following masses were recorded from an unknown substance that contain some amount of KHP in three experimental solutions 0.44, 0.437, and 0.437, if the volume of NaOH IN THE exp are 5, 10, and 8 respectively in order to the masses of unknown substances.1. Find the mass of KHP of each experiment 2.Find % KHP 3. Find average % KHP 4.find standard deviation.( Note; in a previous three 100% KHP experiments the average molarity of NaOH in each experiment are shown below average molarity=5.30x 10^-4+ 4.137x10^-4 + 5.0x10^-4 divided by 3 = 4.81x10^-4 ).

In a titration, the following masses were recorded from an unknown substance that contain some amount of KHP in three experimental solutions 0.44, 0.437, and 0.437, if the volume of NaOH IN THE exp are 5, 10, and 8 respectively in order to the masses of unknown substances.1. Find the mass of KHP of each experiment 2.Find % KHP 3. Find average % KHP 4.find standard deviation.( Note; in a previous three 100% KHP experiments the average molarity of NaOH in each experiment are shown below average molarity=5.30x 10^-4+ 4.137x10^-4 + 5.0x10^-4 divided by 3 = 4.81x10^-4 ).

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.18QAP

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Find the standard deviation (if the mass of KHP in three experimental solutions are 0.467,0.48, and 0.48 and the volume of NaOH IN THE exp are 22,29,24 respectively in order to the masses of KHP ) the average molarity of NaOH in each experiment = 5.30x 10^-4 + 4.137x10^-4 + 5.0x10^-4 divided by 3 = 4.81x10^-4 ).
help with Part B Please. 5. Compare the values of Keq for each of the five solutions: a. Are any of the values outside the sample standard deviation for the set? b. Does the value of the equilibrium constant change when the concentration of either the reactants or products is altered? Why or why not? Explain this based on your calculations of Keq: c. What caused variation (three reasons) in your values of Keq? NOTE: the standard deviation is 3.9x102 The procedure is in the images Data: Unknown solution Equilibrium concentration of {FeNCS2+} in unknown solution(M) Equilibrium concentration [Fe3+](M) Equilibrium concentration [SCN-](M) Keq 1 3.7x10-5 9.6x10-4 1.6x10-4 2.4x102 2 6.2x10-5 9.4x10-4 3.4x10-4 1.9x102 3 1.1x10-4 8.9x10-4 4.9x10-4 2.5x102 4 1.1x10-4 8.9x10-4 6.9x10-4 1.8x102 5 1.8x10-4 8.2x10-4 8.2x10-4 2.7x102
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