Question

+-/4 points My Not How many grams of CHCOOM are needed to make 3.1 liters of aqueous solution of pH-3.5730 sure to back correct from equorum molecular acid molarity to starting molecular acid molarity Submit Answer Practice Another Version

Answer 1

PH   = 3.875

-log[H^+]   = 3.875

   [H^+]   = 0.000133M

at equilibrium [H^+]   = [CH3COO^-]   = 0.000133M

           CH3COOH(aq) -------------> CH3COO^- (aq) + H^+(aq)

I          x                                            0                           0

C      - 0.000133                              0.000133               0.000133

E       x-0.000133                             0.000133               0.000133

           Ka    =   [CH3COO^-][H^+]/[CH3COOH]

          1.8*10^-5   = 0.000133*0.000133/(x-0.000133)

          1.8*10^-5*(x-0.000133)   = 0.000133*0.000133

             x = 0.001115

The molarity of CH3COOH = 0.001115M

no of moles of CH3COOH = molarity *volume in L

                                          = 0.001115*3.1    = 0.0034565moles

mass of CH3COOH   = no of moles * gram molar mass

                                 = 0.0034565*60    = 0.2074g >>>>answer