A solution was prepared by dissolving 0.0170 mole of propionic acid and 0.0179 mole of sodium...

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?

Solutions

Expert 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

queen_honey_blossom answered 5 months ago

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