---Please show all work and equations in details of how thesolution was reached.

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/mol

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---