What is the pH of the solution? What is the pH of a 1.00 L solution which initially has 0.045 mol acetic acid and 0.035 mol sodium acetate once 0.006 mols of NaOH has been added. (Ka for acetic acid is 1.8x10-5. Hint: What are the new concentrations of weak acid and weak base?)

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When NaOH has been added in the buffer solution, the pH eventually will change although not dramatically. The added NaOH will consume the acetic acid (`CH3CO_2H` ) in the solution thus producing additional sodium acetate (`CH3CO_2Na` ). The chemical equation for this can be written as:

  `CH3CO_2H + NaOH...

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When NaOH has been added in the buffer solution, the pH eventually will change although not dramatically. The added NaOH will consume the acetic acid (`CH3CO_2H` ) in the solution thus producing additional sodium acetate (`CH3CO_2Na` ). The chemical equation for this can be written as:

  `CH3CO_2H + NaOH -> CH3CO_2Na + H_2O`

        0.045               0.006

       -0.006               0.006        
------------------------------------------------------------------

        0.039                 0                         0.006             0.006

 

`[CH3CO_2H]_(n ew) = (0.039 mol es)/(1L) = 0.039 M`

`[CH3CO_2Na]_(n ew) = ((0.035 + 0.006)mol es)/(1L) = 0.041 M`

 

Now to get the pH, we can use the Henderson-Hasselbalch equation:

`pH = pK_a + log (([A^-])/([HA]))`

OR

`pH = pK_a_(CH3CO_2H) + log (([CH3CO_2Na])/([CH3CO_2H]))`

OR

`pH = -log [K_a_(CH3CO_2H)]+ log (([CH3CO_2Na])/([CH3CO_2H]))`

Finally, substitute the values and solve for the pH.

`pH = -log [1.8x10^-5] + log (([0.041])/([0.039]))`                 

`pH = 4.74473 + 0.021719`

pH = 4.766  

 

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