Titration of battery acid Equation for the reaction: Volume of NAOH used to neut

Question

Titration of battery acid Equation for the reaction: Volume of NAOH used to neutralize 20.00 mL of battery acid: Moles of NaOH needed to neutralise 20.00mL of battery acid: Moles of NaOH needed to neutralise 20.00mL of battery acid Number of moles of battery acid: Concentration of diluted battery acid: Dilution factor of the battery acid: Concentration of undiluted battery acid: (up to 5 marks awarded for accuracy of titration) List 3 sources of errors which may have contributed to your final result - think carefully about each step in the procedure where errors may have been introduced (operator error would not suffice).

Explanation / Answer

Ans. #2. Volume of NaOH used to neutralize 20.0 mL battery acid = 25.2 mL

= 0.0252 L                             ; [1 L = 1000 mL]

#3. Moles of NaOH = Molarity x Volume of NaOH in liters

                                    = 0.3 M x 0.0252 L

                                    = (0.3 mol/ L) x 0.252 L                                            ; [1 M = 1 mol/L]

                                    = 0.0756 mol

#4. The stoichiometry of balanced reaction shows 2 moles of NaOH neutralizes 1 mol H2SO4 (battery acid).

So, moles of H2SO4 = (1/2) x moles of NaOH used to attain endpoint

                                    = (1/2) x 0.0756 mol

                                    = 0.0378 mol

#5. Given, volume of acid = 20.0 mL = 0.020 L

Concentration of diluted acid = Moles of H2SO4 / Volume of acid in liters

                                                = 0.0378 mol / 0.020 L

                                                = 1.89 mol/ L

                                                = 1.89 M

#6. Concentration H2SO4 (98 %) = 18.4 M

Using                          C1V1 = C2V2                        - equation 1

C1= Concentration, and V1= volume of initial solution 1     ;(Concentrated H2SO4)

C2= Concentration, and V2 = Vol. of final solution 2     ;( diluted acid solution)

Putting the values in equation 1-

            18.4 M x V1 = 1.98 M x 20.0 mL

            Or, V1 = (1.98 M x 20.0 mL) x 18.4 M = 2.05 mL

Therefore, 2.05 mL of original 98% H2SO4 is diluted to 20.0 mL to make battery acid.

Now,

Dilution factor = Volume of original acid taken/ Final volume made upto

                        = 2.05 mL / 20.0 mL

                        = 0.1025 : 1

                        = 0.1025 = 1.025 x 10-1

Therefore, dilution factor = 0.1025

#7. Concentration of undiluted acid (original H2SO4 solution) = 18.4 M. Please see the label on the bottle if available in lab.

#8. Conclusion: Three possible sources of error.

As evident from the table, each titration of the same 20.0 mL battery acid consumed different volumes of standard NaOH. That is, the titration endpoint is not concordant. The following reasons might have caused so-

#A. Error in pipetting battery acid: Pipetting acid while transferring the battery acid to titration flask may result non-concordant endpoint because each time slight different volumes of acid is taken for titration. Note that change is volume only by few drops may cause deviation from concordance.

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