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

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

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Answer 1 · 2017-07-19T07:27:50

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

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

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

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

And if we recall... $log(ab) = log a + log b$ . Thus...

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

$= "pH" + log 10$

$= color(blue)("pH" + 1)$

So, the $"pH"$ 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.

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

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