# If the ratio of weak base to conjugate acid is 10:1, to what extent does the "pH" change and does it increase or decrease?

Jul 19, 2017

Suppose the buffer started with a $\text{pH}$ of, well, "$\text{pH}$", with a $\text{pKa}$ of "$\text{pKa}$", with general concentrations of acid $\text{HA}$ and conjugate base ${\text{A}}^{-}$ (or if you prefer, the salt $\text{NaA}$). Then the Henderson-Hasselbalch equation shows:

"pH" = "pKa" + log\frac(["A"^(-)])(["HA"])

Now, if we have $10$ times the ${\text{A}}^{-}$ (noting that no identity of anything changes, so the $\text{pKa}$ of the acid stays the same), we simply get a new $\text{pH}$ of:

color(blue)("pH"') = "pKa" + log\frac(10["A"^(-)])(["HA"])

And if we recall... $\log \left(a b\right) = \log a + \log b$. Thus...

=> "pKa" + log\frac(["A"^(-)])(["HA"]) + log 10

$= \text{pH} + \log 10$

$= \textcolor{b l u e}{\text{pH} + 1}$

So, the $\text{pH}$ increases by $1$, and the solution becomes $10$ times more basic. That SHOULD make sense... we did, after all, make it so we had $10$ times the conjugate base.