# Question a721d

Mar 17, 2017

pH = 1.61151
$O {H}^{-} = 4.08797 \cdot {10}^{-} 13 M$
HF = 0.855538M
${H}^{+} = 0.024462 M$
${F}^{-} = 0.024462 M$

#### Explanation:

$H F + {H}_{2} O = {H}_{3} {O}^{+} + {F}^{-}$

We can find the concentration of ${H}^{+} \mathmr{and} {H}_{3} {O}^{+}$ by three ways

One is by the ICE table (but this is a 5% rule) and the other is square root which is absolutely correct and the other is Ostwald's law of dillution

Let's set up an ICE table.

$\textcolor{w h i t e}{m m m m m m m m} \text{HF" + "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(m) + color(white)(ml)"F"^"-}$
$\text{I/mol.L"^"-1} : \textcolor{w h i t e}{m m m l} 0.88 \textcolor{w h i t e}{m m m m m} 0 \textcolor{w h i t e}{m m m m m l} 0$
$\text{C/mol.L"^"-1":color(white)(mmml)"-x"color(white)(mmmmll)"+x"color(white)(mlmmml)"+x}$
$\text{E/mol.L"^"-1":color(white)(mmm) 0.88"- x"color(white)(mmml)"x"color(white)(mmmmmll)"x}$

Ka = ${x}^{2} / \left(0.88 - x\right)$

You can ignore x

Ka = x^2/(0.88

$6.8 \times {10}^{-} 4 = {x}^{2} / \left(0.88 - x\right)$

Step 1: Multiply both sides by -x+0.88.
−0.00068x+0.000598=x^2
−0.00068x+0.000598−x^2=x^2−x^2(Subtract ${x}^{2}$ from both sides)
−x^2−0.00068x+0.000598=0

x= {0.0006799999999999999±sqrt(−0.0006799999999999999)^2−4(−1)(0.0005984)}/{2(−1)}

x = 0.02412457847582909 This is the${H}_{3} {O}^{+}$

Another way I told you is very simple

$\sqrt{K a \cdot M} = {H}_{3} {O}^{+}$

$\sqrt{6.8 \cdot {10}^{-} 4 \cdot 0.88 M} = 0.024462$

You can see that -x matters so much

Therefore the concentration of HF

0.88M - x(you remember, we did this in the ice table above) = 0.88M - 0.024462M = 0.855538M

You can see that -x matters so much

Third way is

Ostwald's law of dillution

degree of ionization or $\alpha$ = $\sqrt{\frac{\text{Ka}}{C}}$

${H}^{+} = \alpha \cdot C$

Plug in the variables

$\sqrt{\frac{00068}{0.88}} = 0.02780$

$0.02780 \cdot 0.88 = 0.024462 M$

Now to calculate $O {H}^{-}$

$2 {H}_{2} O = {H}_{3} O + O {H}^{-}$

You know that for any substance in water at 298K the
H_3O^+M * OH^-)M = 1*10^-14

$0.024462 M \cdot \left(O {H}^{-}\right) = 1 \cdot {10}^{-} 14 = 4.08797E-13 M$

$4.08797E-13 = 4.08797 \cdot {10}^{-} 13 M$

Note there are two ways to solve pH . One way is that

$- \log \left(0.024462\right) = p H = 1.61151$

$- \log \left(4.08797 \cdot {10}^{-} 13\right) = p O H = 12.38849$

And then you can subtract the pOH from 14

And if you add these you get a perfect 14

Mar 17, 2017

["H"_3"O"^"+"] = "0.024 mol/L"
$\text{[F]"^"-"color(white)(mm) = "0.024 mol/L}$
["OH"^"-"] color(white)(m)= 4.1 × 10^"-13" color(white)(l)"mol/L"

#### Explanation:

Method 1. Using the 5 % approximation

Let's set up an ICE table.

$\textcolor{w h i t e}{m m m m m m m m} \text{HF" + "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(m) + color(white)(ml)"F"^"-}$
$\text{I/mol.L"^"-1} : \textcolor{w h i t e}{m m l} 0.88 \textcolor{w h i t e}{m m m m m m l} 0 \textcolor{w h i t e}{m m m m m l l} 0$
$\text{C/mol.L"^"-1":color(white)(mmll)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mlmmml)"+} x$
$\text{E/mol.L"^"-1":color(white)(ml) "0.88 -} \textcolor{w h i t e}{l} x \textcolor{w h i t e}{m m m m m l} x \textcolor{w h i t e}{m m x m m m} x$

K_text(a) = x^2/(0.88-x) = 6.8 × 10^"-4"

Test for negligibility

0.88/(6.8 × 10^"-4") = 1300 > 400

$x$ is less than 5 % of the initial concentration of $\left[\text{HF}\right]$.

Since x ≪ 0.88, it can be ignored in comparison with 0.88.

Then

x^2/(0.88) = 6.8 × 10^"-4"

x^2 = 0.88 × 6.8 × 10^"-4"= 5.98 × 10^"-4"

x = 2.45 × 10^"-2"

["H"_3"O"^"+"] = x color(white)(l)"mol/L" = "0.024 mol/L"

["F"^"-"] = x color(white)(l)"mol/L" = "0.024 mol/L"

["OH"^"-"] = K_text(w)/(["H"_3"O"^"+"]) = (1.00 × 10^"-14")/(2.45 ×10^"-2") color(white)(l)"mol/L" = 4.1 × 10^"-13" color(white)(l)"mol/L"

${K}_{\textrm{a}} = {x}^{2} / \text{0.88-x" = 6.8 × 10^"-4}$

x^2 = 6.8 × 10^"-4"("0.88 -"x) = 5.98 × 10^"-4" - 6.8 × 10^"-4"x

x^2 + 6.8 × 10^"-4"x- 5.98 × 10^"-4" = 0

x = 2.41 × 10^"-2"

["H"_3"O"^"+"] = x color(white)(l)"mol/L" = "0.024 mol/L"

["F"^"-"] = x color(white)(l)"mol/L" = "0.024 mol/L"

["OH"^"-"] = K_text(w)/(["H"_3"O"^"+"]) = (1.00 × 10^"-14")/(2.41 × 10^"-2") color(white)(l)"mol/L" = 4.1 × 10^"-13" color(white)(l)"mol/L"#

Conclusion

Within the limitations of the allowed significant figures, the approximate method gives the same result as solving the quadratic.