Question

Consider a solution initially containing 0.400 mol fluoride anion and 0.300 mol of hydrogen fluoride. How many moles of hydrogen fluoride are present after addition of 70.0 ml of 0.600 M HCL to this solution? Answer should be: 0.358 mol fluoride, 0.342 mol HF

Answer 1

no of moles of F^-    = 0.4moles

no of moles of HF   = 0.3 moles

By the addtion of 70ml of 0.6M HCl

no of moles of HCl   = molarity * volume in L

                                  = 0.6*0.07   = 0.042 moles

no of moles of F^- after addition of HCl   = 0.4-0.042   = 0.358 moles

no of moles of HF after addition of HCl   = 0.3+0.042   = 0.342 moles