# Question 99405

Apr 12, 2015

Since you're dealing with a buffer solution, you can use the Henderson-Hasselbalch equation to solve for the pH of the solution.

Now, since you mistyped the volume of the solution, I'll assume it to be 0.100 L - you can use whatever value you had.

So, according to the Henderson-Hasselbalch equation,

$p {H}_{\text{sol}} = p {K}_{a} + \log \left(\frac{\left[{F}^{-}\right]}{\left[H F\right]}\right)$, $\textcolor{b l u e}{\left(1\right)}$

You also need the acid dissociation constant for $H F$, which is listed as being ${K}_{a} = 6.7 \cdot {10}^{- 4}$.

The $p {K}_{a}$ will be

$p {K}_{a} = - \log \left({K}_{a}\right) = - \log \left(6.7 \cdot {10}^{- 4}\right) = 3.17$

Plug your values into equation $\textcolor{b l u e}{\left(1\right)}$ to get the pH

pH_"sol" = 3.17 + log("0.50 M"/"0.25 M") = 3.17 + log(2) = color(green)(3.47)

Now you start adding stuff to your buffer. When you add nitric acid, $H N {O}_{3}$, which is a strong acid, it will react with the conjugate base of your weak acid, ${F}^{-}$.

As a result, you'll have less $N a F$ in the solution, which implies that more $H F$ will be produced.

$H N {O}_{3 \left(a q\right)} + N a {F}_{\left(a q\right)} \to H {F}_{\left(a q\right)} + N a N {O}_{3 \left(a q\right)}$

Now, you need to calculate how many moles of $N a F$ and of $H F$ you initially had in the buffer. To do this, use the given volume and molarity

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{N a F} = \text{0.50 M" * "0.100 L" = "0.050 moles NaF}$

${n}_{H F} = \text{0.25 M" * "0.100 L" = "0.025 moles HF}$

$\text{ } H N {O}_{3 \left(a q\right)} + N a {F}_{\left(a q\right)} \to H {F}_{\left(a q\right)} + N a N {O}_{3 \left(a q\right)}$
I.......0.002...............0.05..............0.025
C.....(-0.002)...........(-0.002).......(+0.002)
E.......0......................0.048............0.027

All the moles of $H N {O}_{3}$ will be consumed; at the same time, 0.002 moles of $H F$ will be produced and the number of moles of $N a F$ will decrease by 0.002.

Calculate the new molarities of the species present in the buffer

$\left[H F\right] = \text{0.027 moles"/"0.100 L" = "0.27 M}$

$\left[{F}^{-}\right] = \text{0.048 moles"/"0.100 L" = "0.48 M}$

Use equation $\textcolor{b l u e}{\left(1\right)}$ to determine the new pH

pH_"sol 2" = 3.17 + log("0.48 M"/"0.27 M") = color(green)(3.42)

Now you add potassium hydroxide, $K O H$ - a strong base. It will react with the weak acid to produce ${F}^{-}$.

$K O {H}_{\left(a q\right)} + H {F}_{\left(a q\right)} \to K {F}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

SIDE NOTE You can ignore the ${K}^{+}$ cation altogether, the important thing in this reaction is ${F}^{-}$.

Once again, set up the ICE table

$\text{ } O {H}_{\left(a q\right)}^{-} + H {F}_{\left(a q\right)} \to {F}_{\left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)}$
I......0.004...........0.027............0.048
C...(-0.004).......(-0.004).........(+0.004)
E.........0...............0.023..............0.052

$\left[H F\right] = \text{0.023 moles"/"0.100 L" = "0.23 M}$

$\left[{F}^{-}\right] = \text{0.052 moles"/"0.100 L" = "0.52 M}$

Finally, use $\textcolor{b l u e}{\left(1\right)}$ to solve for the new pH

pH_"sol 3" = 3.17 + ("0.52 M"/"0.23 M") = color(green)(3.52)#

Adding a strong base increased the pH slightly.