In: Chemistry
A solution was prepared by dissolving 0.0170 mole of propionic acid and 0.0179 mole of sodium propionate in 1.00 L?
What would be the pH of the solution in beaker after 2.00 mL of 0.0154 M HCl were added to 10.0 mL of the prepared solution?
molarity of CH3CH2COOH = no of moles/volume in L
= 0.0170/1 = 0.017M
molarity of CH3CH2COONa = no of moles/volume in L
= 0.0179/1 = 0.0179M
no of moles of HCl = molarity*volume in L
= 0.0154*0.002 = 3.08*10^-5 moles
no of moles of CH3CH2COOH = molarity * volume in L
= 0.017*0.01 = 0.00017moles
no of moles of CH3CH2COONa = molarity * volume in L
= 0.0179*0.01 = 0.000179moles
no of moles of CH3CH2COOH after addition of 3.08*10^-5 moles of HCl = 0.00017+3.08*10^-5 = 0.0002008moles
no of moles of CH3CH2COONa after addition of 3.08*10^-5 moles of HCl = 0.0179-3.08*10^-5 = 0.0178692moles
Pka = 4.87
PH = PKa + log[CH3CH2COONa]/[CH3CH2COOH]
= 4.87 + log0.017892/0.0002008
= 4.87 + 1.9498
= 6.8198 >>>>answer