Question

Calculate `pH` of buffer solution with `0.1` `M` each of `CH_3COOH` & `CH_3COONa`. What will be the change in `pH` on addition of the following: ? (`pK_a` = `1.8 xx 10^-5`). The total volume of solution is `1` `L`.

(i) `0.01` `M` `HCl`
(ii) `0.01` `M` `NaOH`

Answer 1

The buffer equation holds that.........

`pH=pK_a+log_10{([A^-])/([HA])}`

Explanation:

`pH=pK_a+log_10{([A^-])/([HA])}`

Where `HA` is the parent acid; and the `[A^-]` is its conjugate. For the derivation of this formula, which an undergraduate should be able to perform, see here.

And thus for this problem..........

`pH=4.76+log_10{([AcO^-])/([HOAc])}`

And note that `4.76=-log_10(1.8xx10^-5)`

Now the starting conditions specify that there are equal concentrations of acetate and free acid, and thus `pH=pK_a=4.76`. Why? Because `log_(10)1=0`, so given the equation, `[AcO^-]=[HOAc]`, and `([AcO^-])/([HOAc])=1`.

Anyway, work thru the problems, and post them back in this thread. The `pH` will be round about `4.76`, i.e. the `pK_a` of the parent acid.