A solution of sodium hydroxide (NaOH) was standardized against potassium hydroge

Question

A solution of sodium hydroxide (NaOH) was standardized against potassium hydrogen phthalate (KHP) A Known mass of KHP was titrated with the NaOH solution until a light pink color appeared using phenolphthalein indicator. Using the volume of NaOH required to neutralize KHP and the number of moles of KHP titrated, the concentration of the NaOH solution was calculated Molecular formula of Potassium hydrogen phthalate HKC8H4O4 Mass of KHP used for standardization (g) 0 5147 Volume of NaOH required to neutralize KHP (mL) 13 03 Calculate the concentration of NaOH solution in (mol/L) Answer: A vinegar (acetic acid) solution of unknown concentration was titrated to the light pink endpoint with the standardized NaOH solution The weight Volume % of the vinegar solution were calculated Molecular formula of Acetic acid C2H4O2 Volume of vinegar sample titrated (mL) 5 00 Volume of NaOH required to neutralize vinegar in (mL) 8 90 Concentration of NaOH in mol/L. 0 1983 Calculate the weight/volume percentage of the vinegar solution (g/100 ml) Answer: A vitamin C (ascorbic acid) tablet was dissolved in approximately 50 mL of distilled water and titrated with the standardized NaOH solution. From the results of this titration, the mg of ascorbic acid m the tablet was calculated

Explanation / Answer

           HKC8H4O4 + NaOH --------> NaC8H4O4 + H2O

no of moles of HKC8H4O4 = W/G.M.Wt

                                           = 0.5147/204.2

                                            = 0.00252 moles

from balanced equation

1 mole of HKC8H4O4 react with 1 mole of NaOH

0.00252 moles of HKC8H4O4 react with = 1*0.00252/1 = 0.00252 moles of NaOH

no of moles of NaOH = molarity * volume in L

           0.00252          = molarity*0.01303

           molarity     = 0.00252/0.01303   = 0.193M >>>> answer

         CH3COOH + NaOH -------> CH3COONa + H2O

         1 mole         1mole

       no of moles of CH3COOH = no of moles of NaOH

         no of moles of NaOH = molarity * volume in L

                                                    = 0.1983*0.0089

                                                   = 0.00176 moles

              no of moles of CH3COOH = 0.00176moles

                  molarity of CH3COOH = 0.00176/0.005 = 0.352 M

                     mass of CH3COOH = no of mole* gram molar mass

                                                     = 0.00176*60 = 0.1056g

                    weight volume of CH3COOH = W*100/v

                                                                 = 0.1056*100/5

                                                                 =2.112%

                                                      

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