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A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M...

A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate use 1.80 *10^-5 as Ka for acetic acid.

a.) what is the pH of the solution

b.) Is the solution acidic or basic

c.) What is the pH of a solution that results when 3.00ml of 0.034 M HCl is added to 0.200L of the original buffer

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Expert Solution

PKa   = -logKa

           = -log(1.8*10^-5)

             = 4.75

PH   = PKa + log[CH3COONa]/[CH3COOH]

       = 4.75 + log0.6/0.2

       = 4.75 + 0.477

       = 5.227

b. PH = 5.227

The solution is acidic

c.

no of moles of CH3COONa = molarity *volume in L

                                         = 0.6*0.2   = 0.12moles

no of moles of CH3COOH = molarity *volume in L

                                        = 0.2*0.2   = 0.04moles

no of moles of HCl   = molarity * volume in L

                              = 0.034*0.003   = 0.000102moles

no of moles of CH2COONa after addition o f0.000102moles of HCl   = 0.12-0.000102   = 0.119898moles

no of moles of CH2COOH after addition o f 0.000102moles of HCl    = 0.04+0.000102   = 0.040102moles

PH   = PKa + log[CH3COONa]/[CH3COOH]

       = 4.75 + log0.119898/0.040102

       = 4.75 + 0.4756

       = 5.2256 >>>>answer

queen_honey_blossom answered 1 day ago
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