What is the concentration of acetate ions @ the stoichiometric point in the titration of 0.018 M CH3COOH(aq) with 0.036 M NaOH(aq)? For acetic acid, Ka = 1.8×10−5

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`CH_3COOH+NaOH rarr CH_3COONa+H_2O`

 

Let us say we have 1L of `CH_3COOH ` solution.

According to the stoichiometry;

`CH_3COOH:NaOH = 1:1`

`CH_3COOH:CH_3COONa = 1:1`

 

Since we are adding enough NaOH full reaction will be occurred. We can assume that there is no partial reactions.

CH_3COOH moles present before tritration `= 0.018`

NaOH moles required for titration `= 0.018`

Required NaOH volume from 0.036M `= 1/0.036*0.018 = 0.5`

 

`CH_3COOH:CH_3COONa = 1:1`

Final CH_3COONa mole in the solution `= 0.018`

Final volume of the solution` = 1+0.5 = 1.5L`

 

`CH_3COONa rarr CH_3COO^-+Na^+`

 

`[CH_3COO^-] = 0.018/1.5 = 0.012M`

 

So the concentration of `CH_3COO^-` ions is 0.012M.

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