To make a solution with a pH = 4.20 a student used thefollowing procedure : a ce

Question

To make a solution with a pH = 4.20 a student used thefollowing procedure :
a certain amount of sodium acetate along with 0.250 moles of aceticacid is added to enough water to make a solution of 1.00 L

How many grams of sodiumacetate were added ? Ka = 1.80 x 10-5 To make a solution with a pH = 4.20 a student used thefollowing procedure :
a certain amount of sodium acetate along with 0.250 moles of aceticacid is added to enough water to make a solution of 1.00 L

How many grams of sodiumacetate were added ?

Explanation / Answer

    The solution of acetic acid and sodiumacetate form an acidic buffer .              pH = pKa + log [salt] / [acid]            [ CH3COOH] = 0.250 / 1.0 L                                     =0.250 M                      given Ka = 1.8 x 10-5               pKa = 4.744                4.20= 4.744 + log [ CH3COONa ] /[CH3COOH]                      = 4.744 + log [CH3COONa] / 0.250           [CH3COONa ] = 0.0713 M      molar mass of CH3COONa =82.3 g/mol                0.0713 = mass of (CH3COONa / 82.3 ) / 1L            mass of CH3COONa = 5.869 g

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