---Please show all work and equations in details of how thesolution was reached.
ID: 686450 • Letter: #
Question
---Please show all work and equations in details of how thesolution was reached....thanks---3) A 0.2678g sample of an unknown acid requires 27.21 mL of0.1164 M NaOH for neutralization to a phenolphalein endpoint. There are 0.35 mL of 0.1012 M HCl used forback-titration.
a. How many moles of OH- are used? How many moles of H+ from HCl?
___________moles OH- ____________molesH+
b. How many moles of H+ are there in thesolid acid?
____________moles H+ in solid
c. What is the molar mass of the unknownacid?
_______________ g
Explanation / Answer
(a) Moles of OH- used = moles of NaOH = molarity *volume / 1000 = 0.1164 M * 27.21 mL / 1000 = 0.003167 moles . Moles of H+ = moles of HCl = molarity * volume /1000 = 0.1012 M * 0.35 mL / 1000 = 3.54 * 10-5 moles . (b) Moles of H+ in acid = moles of OH-used for neutralization. Moles of OH- used = moles of OH- used initially -moles of H+ in reverse titration =0.003167 - (3.54 * 10-5 ) =0.003131 moles =moles of H+ in the solid acid . (c) Assuming it is a mono protic acid, molesof H+ in the acid is equal to moles of acid. Therefore Moles of acid = 0.003131 moles mass = 0.2678 g Molar mass = mass/ moles = 0.2678 g / 0.003131 moles = 85.53 g/molRelated Questions
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