# Question bf279

Dec 6, 2016

$\text{pH} = 5.7$

#### Explanation:

Your ultimate goal here is to figure out how much weak acid, $\text{HA}$, and how much conjugate base, ${\text{A}}^{-}$, you have in the target solution.

You already know that both buffers have

["HA"] = "0.5 M"

so your starting point here will be to figure out the concentration of ${\text{A}}^{-}$ in buffer $\text{X}$ and in buffer $\text{Y}$.

To do that, use the Henderson - Hasselbalch equation, which for a weak acid - conjugate base buffer takes the form

color(blue)(ul(color(black)("pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))))

Calculate the $\text{p} {K}_{a}$ of the acid by using

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{p} {K}_{a} = - \log \left({K}_{a}\right)}}}$

In your case, you will have

$\text{p} {K}_{a} = - \log \left(1.0 \cdot {10}^{- 5}\right) = 5.0$

Now, calculate the concentration of ${\text{A}}^{-}$ in the two buffers

$\textcolor{w h i t e}{a}$

• $\underline{\text{For buffer X}}$

This buffer has $\text{pH} = 4.0$, which means that you get

$4.0 = 5.0 + \log \left(\left(\left[\text{A"^(-)]_"X")/(["HA}\right]\right)\right)$

Rearrange to solve for ["A"^(-)]_"X"/(["HA"])

log(["A"^(-)]_"X"/(["HA"])) = -1.0

10^log(["A"^(-)]_"X"/(["HA"])) = 10^(-1.0)

This will get you

["A"^(-)]_"X"/(["HA"]) = 0.10

which results in

$\left[\text{A"^(-)]_"X" = 0.10 * ["HA}\right]$

["A"^(-)]_"X" = 0.10 * "0.5 M" = "0.050 M" " "color(orange)((1))

$\textcolor{w h i t e}{a}$

• $\underline{\text{For buffer Y}}$

This buffer has $\text{pH} = 6.0$, which means that you get

$6.0 = 5.0 + \log \left(\left(\left[\text{A"^(-)]_"Y")/(["HA}\right]\right)\right)$

Rearrange to solve for ["A"^(-)]_"Y"/(["HA"])

log(["A"^(-)]_"Y"/(["HA"])) = 1.0

This will get you

["A"^(-)]_"Y"/(["HA"]) = 10

which results in

$\left[\text{A"^(-)]_"Y" = 10 * ["HA}\right]$

["A"^(-)]_"Y" = 10 * "0.5 M" = "5.0 M" " "color(orange)((2))

Now, you are told that you must mix equal volumes of buffer $\text{X}$ and of buffer $\text{Y}$. Right from the start, you could say that because the volume doubles, the concentration of the weak acid remain unchanged. $\textcolor{p u r p \le}{\left(\text{*}\right)}$

That is the case because you're essentially doubling the number of moles of weak acid and the volume of the solution.

If you take $x$ $\text{L}$ to be the volume of the two buffers, and using $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ and $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$, you can say that buffer $\text{X}$ will contain

x color(red)(cancel(color(black)("L buffer"))) * "0.5 moles HA"/(1color(red)(cancel(color(black)("L buffer")))) = (0.5 * x)" moles HA"

x color(red)(cancel(color(black)("L buffer"))) * "0.050 moles A"^(-)/(1color(red)(cancel(color(black)("L buffer")))) = (0.050 * x)" moles A"^(-)

Similarly, buffer $\text{Y}$ will contain

x color(red)(cancel(color(black)("L buffer"))) * "0.5 moles HA"/(1color(red)(cancel(color(black)("L buffer")))) = (0.5 * x)" moles HA"

x color(red)(cancel(color(black)("L buffer"))) * "5.0 moles A"^(-)/(1color(red)(cancel(color(black)("L buffer")))) = (5.0 * x)" moles A"^(-1)

The resulting solution, which has a volume of

$x \textcolor{w h i t e}{.} \text{L" + xcolor(white)(.)"L" = 2xcolor(white)(.)"L}$

will contain

$\left\{\begin{matrix}\left(0.5 x\right) {\text{ moles" + (0.5x)" moles" = x" moles HA" \\ (0.050x)" moles" + (5.0x)" moles" = 5.05x" moles A}}^{-}\end{matrix}\right.$

The concentrations of the two species in the resulting solution will be

["HA"] = (color(red)(cancel(color(black)(x)))"moles")/(2color(red)(cancel(color(black)(x)))"L") = "0.5 M" -> the concentration remained unchanged, as predicted in $\textcolor{p u r p \le}{\left(\text{*}\right)}$

["A"^(-)] = (5.05color(red)(cancel(color(black)(x)))"moles")/(2color(red)(cancel(color(black)(x)))"L") = "2.525 M"

Finally, use the Henderson - Hasselbalch equation to find the $\text{pH}$ of the resulting solution

"pH" = 5.0 + log( (2.525 color(red)(cancel(color(black)("M"))))/(0.5color(red)(cancel(color(black)("M"))))) = color(darkgreen)(ul(color(black)(5.7)))#

The answer is rounded to one decimal place, since that is how many sig figs you have for the concentration of the weak acid.