# 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?)

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