Please Report answer to the correct sig figs for all questions Consider again th
ID: 877334 • Letter: P
Question
Please Report answer to the correct sig figs for all questions
Consider again the titration of HC2H3O2 with NaOH. Let’s say that you need to add 0.0269 moles of NaOH in order for the indicator dye to change color. How many moles of HC2H3O2 must you have had to start?
Consider the titration of HC2H3O2 with NaOH. If it requires 0.00385 mol of NaOH to reach the endpoint, and if we had originally placed 13.71 mL of HC2H3O2 in the Erlenmeyer flask to be analyzed, what is the molarity of the original HC2H3O2 solution?
Consider the titration of H3C6H5O7 with NaOH. Let’s say that you need to add 0.035 moles of NaOH in order for the indicator dye to change color. How many moles of H3C6H5O7 must you have had to start?
Explanation / Answer
Consider again the titration of HC2H3O2 with NaOH. Let’s say that you need to add 0.0269 moles of NaOH in order for the indicator dye to change color. How many moles of HC2H3O2 must you have had to start?
Solution:- balanced reaction equation for the acetic and sodium hydroxide is as follows
NaOH + HC2H3O2----- > H2O + HC2H3Ona
According to the balanced reaction equation the moles ratio of the acetic acid and sodium hydroxide is 1:1
Therefore the moles of the acetic acid needed for the titration with the sodium hydroxide are same as moles of the sodium hydroxide used
Since we used 0.0269 moles of NaOH therefore we need to start with the 0.0269 moles of HC2H3O2
Consider the titration of HC2H3O2 with NaOH. If it requires 0.00385 mol of NaOH to reach the endpoint, and if we had originally placed 13.71 mL of HC2H3O2 in the Erlenmeyer flask to be analyzed, what is the molarity of the original HC2H3O2 solution?
Solution :- since mole ratio of the HC2H3O2 and NaOH is 1:1 therefore the moles of the HC2H3O2 used are same as moles of the NaOH
Therefore miles of the HC2H3O2 = 0.00385 mol
Volume of the HC2H3O2= 13.71 ml = 0.01371 L
Now lets calculate the molarity of the acid using the moles and volume
Molarity = moles / volume in liter
Lets put the values in the formula
Molarity of the HC2H3O2 = 0.00385 mol / 0.01371 L
=0.281 M
Molarity of the original solution HC2H3O2 = 0.281 M
Consider the titration of H3C6H5O7 with NaOH. Let’s say that you need to add 0.035 moles of NaOH in order for the indicator dye to change color. How many moles of H3C6H5O7 must you have had to start?
Solution :- Given data , 0.035 moles of NaOH
Moles of H3C6H5O7= ?
balanced reaction equation is as follows
H3C6H5O7 + 3NaOH ----- > Na3C6H5O7 + 3 H2O
According to the balanced reaction equation mole ratio of the H3C6H5O7 and NaOH is 1 : 3
Using this moles ratio lets calculate the moles of the H3C6H5O7 needed to react with 0.035 moles of NaOH
(0.035 mol NaOH * 1 mol H3C6H5O7)/ 3 mol NaOH = 0.0117 mol H3C6H5O7
Therefore moles of H3C6H5O7 needed = 0.0117 moles.