Question

0.2m of benzoic acid and 0.10m benzoate are dissolved in 1L of pure water. What's the pH of this solution given that ka of benzoic acid is 6.3×10-5?

Answer 1

`pH=3.90`

Explanation:

What you have described here is a buffer equation, for which...

`pH=underbrace(-log_10K_a)_(pK_a)+log_10{([C_6H_5CO_2^(-)])/([C_6H_5CO_2H])}`...

And so...`pH=underbrace(-log_10(6.30xx10^-5))_"4.20"+underbrace(log_10((0.10*mol*L^-1)/(0.20*mol*L^-1)))_"-0.301"`

`=4.20-0.301=3.90`

The `pH` moves slightly in the direction FROM the `pK_a` where the acid or base is in excess. And here, because the acid is in excess, the `pH` moves in the negative direction with respect to `pK_a`. What would the `pH` be in the scenario where `[C_6H_5CO_2^(-)]-=[C_6H_5CO_2H]`?