Derivation

In this problem, we are trying to solve for a variable that is inside of a logarithm. There are several steps, but we will ultimately need to use a process called exponentiation.

That is:

${10}^{log(x)}=x$

At some point in the derivation, you will need to exponentiate both sides of the equation.

The Henderson$-$Hasselbalch equation describes the $pH$ of a solution under buffering conditions:

$pH=p{K}_a+log(\frac{[A^{-}]}{[HA]})$

where $pH=-log([H^{+}{]}_{eq})$ and $p{K}_a=-log{K}_a*[A^{-}]$and $HA$ are the molar concentrations of the basic and acidic components of the buffer, respectively. It is often of

practical importance to re$-$arrange the Henderson$-$Hasselbalch equation to express the ratio of bae$-$to$-$acid, $\frac{[A^{-}]}{[HA]},$ in terms of the pH$.$ This allows scientists to determine the

appropriate base$-$to$-$acid ratio necessary to reach a desired $pH.$

Derive an expression for $\frac{[A^{-}]}{[HA]}$ in terms of $pH$ and $p{K}_a,$ and select an equivalent equation from the list below. Be careful with negative signs!

$\frac{[A^{-}]}{[HA]}=log(p{K}_a-pH)$

$\frac{[A^{-}]}{[HA]}=log(pH-p{K}_a)$

$\frac{[A^{-}]}{[HA]}=10*(p{K}_a-pH)$

$\frac{[A^{-}]}{[HA]}={10}^{(pH-p{K}_a)}$

$\frac{[A^{-}]}{[HA]}=10*(pH-p{K}_a)$

$\frac{[A^{-}]}{[HA]}=\frac{p{K}_a-pH}{log}$

$\frac{[A^{-}]}{[HA]}={10}^{(p{K}_a-pH)}$

$\frac{[A^{-}]}{[HA]}=\frac{pH-p{K}_a}{log}$