# A buffer solution was prepared from 0.615*L of weak acid HA at 0.250*mol*L^-1 concentration, and a 0.500*L volume of NaOH at 0.130*mol*L^-1 concentration. What is the resultant pH of this buffer?

Jul 2, 2017

See here for background....... We finally get $p H = 4.31$

#### Explanation:

The buffer equation tells us that......

Thus $p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

We gots $p {K}_{a} = - {\log}_{10} \left(3.52 \times {10}^{-} 5\right) = 4.45$, we have to work out the concentrations of free acid, and conjugate base.

Now addition of the strong base to the weak acid.....immediately gives us stoichiometric quantities of $N {a}^{+} {A}^{-}$, where $H A$ was the conjugate acid........

$H A \left(a q\right) + N a O H \left(a q\right) \rightarrow N {a}^{+} {A}^{-} + {H}_{2} O$

Starting moles of ..................$H A = 0.615 \cdot L \times 0.250 \cdot m o l \cdot {L}^{-} 1 = 0.154 \cdot m o l$ $\left(i\right)$

Moles of $N a O H = 0.500 \cdot L \times 0.130 \cdot m o l \cdot {L}^{-} 1 =$

$0.0650 \cdot m o l$.

And so INITIALLY we gots $\left(0.154 - 0.0650\right) \cdot m o l$ $H A$ $=$ $0.089 \cdot m o l$ $\left(i i\right)$

And $0.0650 \cdot m o l$ ${A}^{-}$. These molar quantities are dissolved in a volume of $\left(500 + 615\right) \cdot m L = 1115 \cdot m L = 1.115 \cdot L$, but it doesn't really matter because.....

$p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$, and since in the log quotient....the volume of solution $\text{IS THE SAME}$ we can ignore it, and just use the molar quantities of acid and conjugate base.

And so FINALLY.......

$p H = {\underbrace{- {\log}_{10} \left(3.52 \times {10}^{-} 5\right)}}_{\textcolor{red}{p {K}_{a}}} + {\log}_{10} \left\{\frac{0.0650}{0.089}\right\}$

$=$ $- \left(- 4.45\right) + \left(- 0.136\right) = 4.31 \ldots \ldots$

Now had $\left[{A}^{-}\right] = \left[H A\right]$, then the $p H$ of the resultant buffer would have been equal to $p {K}_{a}$. Why because ${\log}_{10} 1 = 0$, and had we put this value back into the expression we get $p H = p {K}_{a}$.

The final $p H$ is a trifle less than the $p {K}_{a}$, which makes sense given that there were more moles of acid than there were moles of base.....and the $p H$ should be reduced from $p {K}_{a}$ (because more $\left[{H}_{3} {O}^{+}\right]$.

I acknowledge that I have gone on for a long time about nothing to the power of less. But all I have done is to use $\log$ functions and calculated the molar quantities of acid and base, even tho it takes forever. Good luck.