Question #a721d

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Question #a721d

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Answer 1 · 2017-03-17T11:13:11

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

Explanation:

$H F + H_{2} O = H_{3} O^{+} + F^{-}$ $HF + H_2O = H_3O^+ + F^-$

We can find the concentration of $H^{+} or H_{3} O^{+}$ $H^+ or H_3O^+$ 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.

$m m m m m m m m HF + H_{2} O ⇌ H_{3} O^{+} m + m l F^{-}$ $color(white)(mmmmmmmm)"HF" + "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(m) + color(white)(ml)"F"^"-"$
${I/mol.L}^{-1} : m m m l 0.88 m m m m m 0 m m m m m l 0$ $"I/mol.L"^"-1":color(white)(mmml)0.88color(white)(mmmmm)0color(white)(mmmmml)0$
${C/mol.L}^{-1} : m m m l -x m m m m l l +x m l m m m l +x$ $"C/mol.L"^"-1":color(white)(mmml)"-x"color(white)(mmmmll)"+x"color(white)(mlmmml)"+x"$
${E/mol.L}^{-1} : m m m 0.88 - x m m m l x m m m m m l l x$ $"E/mol.L"^"-1":color(white)(mmm) 0.88"- x"color(white)(mmml)"x"color(white)(mmmmmll)"x"$

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

You can ignore x

Ka = $\frac{x^{2}}{0.88}$ $x^2/(0.88$

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

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

x= $\frac{0.0006799999999999999 \pm {\sqrt{- 0.0006799999999999999}}^{2} - 4 (- 1) (0.0005984)}{2 (- 1)}$ ${0.0006799999999999999±sqrt(−0.0006799999999999999)^2−4(−1)(0.0005984)}/{2(−1)}$

x = 0.02412457847582909 This is the $H_{3} O^{+}$ $H_3O^+$

Another way I told you is very simple

$\sqrt{K a \cdot M} = H_{3} O^{+}$ $sqrt(Ka *M) = H_3O^+$

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

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 $α$ $alpha$ = $\sqrt{\frac{Ka}{C}}$ $sqrt("Ka"/C)$

$H^{+} = α \cdot C$ $H^+ = alpha * C$

Plug in the variables

$\sqrt{\frac{00068}{0.88}} = 0.02780$ $sqrt(00068 /0.88) = 0.02780$

$0.02780 \cdot 0.88 = 0.024462 M$ $0.02780 * 0.88 = 0.024462M$

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

$2 H_{2} O = H_{3} O + O H^{-}$ $2H_2O = H_3O + OH^-$

You know that for any substance in water at 298K the
$H_{3} O^{+} M \cdot O H^{-} M = 1 \cdot 10^{- 14}$ $H_3O^+M * OH^-)M = 1*10^-14$

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

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

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

$- log (0.024462) = p H = 1.61151$ $-log( 0.024462) = pH = 1.61151$

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

And then you can subtract the pOH from 14

And if you add these you get a perfect 14

Answer 2 · 2017-03-17T19:32:53

$[H_{3} O^{+}] = 0.024 mol/L$ $["H"_3"O"^"+"] = "0.024 mol/L"$
${[F]}^{-} m m = 0.024 mol/L$ $"[F]"^"-"color(white)(mm) = "0.024 mol/L"$
$[{OH}^{-}] m = 4.1 \times 10^{-13} l mol/L$ $["OH"^"-"] color(white)(m)= 4.1 × 10^"-13" color(white)(l)"mol/L"$

Explanation:

Method 1. Using the 5 % approximation

Let's set up an ICE table.

$m m m m m m m m HF + H_{2} O ⇌ H_{3} O^{+} m + m l F^{-}$ $color(white)(mmmmmmmm)"HF" + "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(m) + color(white)(ml)"F"^"-"$
${I/mol.L}^{-1} : m m l 0.88 m m m m m m l 0 m m m m m l l 0$ $"I/mol.L"^"-1":color(white)(mml)0.88color(white)(mmmmmml)0color(white)(mmmmmll)0$
${C/mol.L}^{-1} : m m l l - x m m m m m m + x m l m m m l + x$ $"C/mol.L"^"-1":color(white)(mmll)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mlmmml)"+"x$
${E/mol.L}^{-1} : m l 0.88 - l x m m m m m l x m m x m m m x$ $"E/mol.L"^"-1":color(white)(ml) "0.88 -"color(white)(l) xcolor(white)(mmmmml)xcolor(white)(mmxmmm)x$

$K_{a} = \frac{x^{2}}{0.88 - x} = 6.8 \times 10^{-4}$ $K_text(a) = x^2/(0.88-x) = 6.8 × 10^"-4"$

Test for negligibility

$\frac{0.88}{6.8 \times 10^{-4}} = 1300 > 400$ $0.88/(6.8 × 10^"-4") = 1300 > 400$

∴ $x$ $x$ is less than 5 % of the initial concentration of $[HF]$ $["HF"]$ .

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

Then

$\frac{x^{2}}{0.88} = 6.8 \times 10^{-4}$ $x^2/(0.88) = 6.8 × 10^"-4"$

$x^{2} = 0.88 \times 6.8 \times 10^{-4} = 5.98 \times 10^{-4}$ $x^2 = 0.88 × 6.8 × 10^"-4"= 5.98 × 10^"-4"$

$x = 2.45 \times 10^{-2}$ $x = 2.45 × 10^"-2"$

$[H_{3} O^{+}] = x l mol/L = 0.024 mol/L$ $["H"_3"O"^"+"] = x color(white)(l)"mol/L" = "0.024 mol/L"$

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

$[{OH}^{-}] = \frac{K_{w}}{[H_{3} O^{+}]} = \frac{1.00 \times 10^{-14}}{2.45 \times 10^{-2}} l mol/L = 4.1 \times 10^{-13} l mol/L$ $["OH"^"-"] = K_text(w)/(["H"_3"O"^"+"]) = (1.00 × 10^"-14")/(2.45 ×10^"-2") color(white)(l)"mol/L" = 4.1 × 10^"-13" color(white)(l)"mol/L"$

Method 2. Solving the quadratic

We had the expression

$K_{a} = \frac{x^{2}}{0.88-x} = 6.8 \times 10^{-4}$ $K_text(a) = x^2/"0.88-x" = 6.8 × 10^"-4"$

$x^{2} = 6.8 \times 10^{-4} (0.88 - x) = 5.98 \times 10^{-4} - 6.8 \times 10^{-4} x$ $x^2 = 6.8 × 10^"-4"("0.88 -"x) = 5.98 × 10^"-4" - 6.8 × 10^"-4"x$

$x^{2} + 6.8 \times 10^{-4} x - 5.98 \times 10^{-4} = 0$ $x^2 + 6.8 × 10^"-4"x- 5.98 × 10^"-4" = 0$

$x = 2.41 \times 10^{-2}$ $x = 2.41 × 10^"-2"$

$[H_{3} O^{+}] = x l mol/L = 0.024 mol/L$ $["H"_3"O"^"+"] = x color(white)(l)"mol/L" = "0.024 mol/L"$

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

$[{OH}^{-}] = \frac{K_{w}}{[H_{3} O^{+}]} = \frac{1.00 \times 10^{-14}}{2.41 \times 10^{-2}} l mol/L = 4.1 \times 10^{-13} l mol/L$ $["OH"^"-"] = K_text(w)/(["H"_3"O"^"+"]) = (1.00 × 10^"-14")/(2.41 × 10^"-2") color(white)(l)"mol/L" = 4.1 × 10^"-13" color(white)(l)"mol/L"$

Conclusion

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

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