10.00 mL of vinegar (mass=10.05 g) requires 14.77 mL of 0.4926 M NaOH to reach the end point. Calculate the molarity and mass percent of the acetic acid in the vinegar.

 Solution:-

Calculation of molarity :-

                  at the end point base  M1*V1*N1 = M2*V2*N2

          molarity of vinegar acid (M1) = to find

                      volume of vinegar V1 = 10 ml 

       It gives one H + ion  so that N1= 1 

                   Molarity of NaOH M2 = 0.4926

                           volume of NaOH = 14.77 ml 

                 It gives one OH- so N2 = 1   

                                                                                    Now use the formula  

                                 M1*V1*N1 = M2*V2*N2 

                                                                                   plug the values in this formula we get 

                                    M1*10*1 = 0.4926*14.77*1 

                                         10 M1 = 7.275702

                                              M1 = 0.7276 M

                                                              Answer molarity of vinegar is 0.7276 M 

Calculation of mass % :-

number of moles of base  = Molarity * volume in liter  of base 

                                       =(0.4926 moles /liter)*(14.77ml /1000)    

                                                                                        divided by 1000 to convert volume ml to  liter

                                       = 0.007275702 moles

reaction between acid and base 

                         CH3COOH + NaOH ---> CH3COONa  + H2O 

                                                                  here one moles of acid is reacting with 1 moles of base 

so number of moles of acid = number of moles of base 

                                         =  0.007275702 moles 

 Molar mass of acetic acid  CH3COOH = 60.05 g / moles

Mass of vinegar is = molar mass * number of moles 

                            =60.05 g/moles  * 0.007275702 moles 

                           =  0.4369059051 gram 

grams % of vinegar =  mass of acetic acid * 100/ total mass vinegar 

                              = 0.4369059051*100/ 10.05 % 

                              = 4.35%

 Answer mass % of vinegar is 4.35%

10.00 mL of vinegar (mass=10.05 g) requires 14.77 mL of 0.4926 M NaOH to reach the end point. Calculate the molarity and mass percent of the acetic acid in the vinegar.