# If you want to prepare a solution that has "pH" = 8.51 at 25^@ "C", what initial concentration of "NaF" would you need to achieve?

Nov 30, 2016

$\textsf{\left[H F\right] = 0.69 \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

$\textsf{{F}^{-}}$ is the co - base of $\textsf{H F}$:

$\textsf{{F}^{-} + {H}_{2} O r i g h t \le f t h a r p \infty n s H F + O {H}^{-}}$

You need to look up $\textsf{p {K}_{b}}$ for the $\textsf{{F}^{-}}$ ion which = 10.82.

From the ICE table you get this expression:

sf(pOH=1/2(pK_a-log[F^-])`

$\textsf{p O H = 14 - p H = 14 - 8.51 = 5.49}$

$\therefore$sf(5.49=1/2(10.82-log[F^-])

$\textsf{\frac{1}{2} \log \left[{F}^{-}\right] = - 0.08}$

$\textsf{\log \left[{F}^{-}\right] = - 0.16}$

$\therefore$$\textsf{\left[{F}^{-}\right] = 0.69 \textcolor{w h i t e}{x} \text{mol/l}}$

Dec 1, 2016

I am also getting approximately $\text{0.70 M}$. Specifically I get $\text{0.708 M}$. You might want to try doing the problem backwards and work from the solved concentration to check that the $\text{pH}$ still turns out to be close to $8.51$.

As I mentioned earlier, you were given the pH which allows you to find the equilibrium concentration of ${\text{H}}^{+}$. Note the main equilibrium in this problem is:

${\text{F"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HF"(aq) + "OH}}^{-} \left(a q\right)$

$\text{I"" "c" M"" "" "" "-" "" "" ""0 M"" "" ""0 M}$
$\text{C"" "-x " ""M"" "-" "" "+x " M"" "+x" M}$
$\text{E"" "(c - x) "M"" "-" "" "" "x" M"" "" "x" M}$

where we let $c$ be the initial concentration of ${\text{F}}^{-}$, which is equal to that of $\text{NaF}$ by their 1:1 stoichiometry.

You should have the ${K}_{a}$ in your book, but I shall use the $\text{pKa}$ of $3.17$ for $\text{HF}$, which corresponds to a $\text{pKb}$ of $10.83$ for ${\text{F}}^{-}$, since $\text{pKw" = "pKa" + "pKb} = 14$ at ${25}^{\circ} \text{C}$.

Therefore, ${K}_{b} = {10}^{- \text{pKb}} = {10}^{- 10.83} = 1.479 \times {10}^{- 11}$. So, the ${K}_{a}$ in your book should be around $6.671 \times {10}^{- 4}$. Now, using the definition of ${K}_{b}$:

${K}_{b} = \left(\left[{\text{HF"]["OH"^(-)])/(["F}}^{-}\right]\right) = {x}^{2} / \left(c - x\right)$

Since we were given that $\text{pH} = 8.51$, we know that $\text{pOH" = 14 - "pH} = 5.49$ at ${25}^{\circ} \text{C}$. Therefore:

$\left[{\text{OH}}^{-}\right]$ at equilibrium is ${10}^{- \text{pOH}} = 3.236 \times {10}^{- 6}$ $\text{M} = x$.

This means we already solved for $x$. So, we can plug it back into ${K}_{b}$:

${K}_{b} = \left(3.236 \times {10}^{- 6} \text{M")^2/(c - 3.236xx10^(-6) "M}\right) = 1.479 \times {10}^{- 11}$

By the small ${K}_{b}$ approximation, since ${K}_{b} \text{<<} {10}^{- 5}$, we can say that the true answer allows us to neglect the $- 3.236 \times {10}^{- 6} \text{M}$ term that is being subtracted in the denominator with minimal error.

As a result, our evaluation simplifies to (where ${K}_{b}$ implicitly has units of $\text{M}$ since it had units of $\text{M"^2/"M}$ from its definition):

${\left(1.479 \times {10}^{- 11} \text{M")c = (3.326xx10^(-6) "M}\right)}^{2}$

=> color(blue)(c = ["F"^(-)] = ["NaF"] = "0.708 M")