Question #721ab

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Question #721ab

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Answer 1 · 2017-03-14T02:18:18

pH = 1.81

Explanation:

Let's set up an ICE table for the calculation of $[H_{3} O^{+}]$ $["H"_3"O"^+]$ .

$m m m m m m m HF l + l H_{2} O l ⇌ l H_{3} O^{+} l + l F^{-}$ $color(white)(mmmmmmm)"HF"color(white)(l) +color(white)(l) "H"_2"O"color(white)(l) ⇌ color(white)(l) "H"_3"O"^+color(white)(l) +color(white)(l) "F"^"-"$
${I/mol\cdotL}^{-1} : m l l 0.35 m m m m m m m l l 0 m m m l l 0$ $"I/mol·L"^"-1": color(white)(mll)0.35color(white)(mmmmmmmll)0color(white)(mmmll)0$
${C/mol\cdotL}^{-1} : m m - x m m m m m m m l +x m m m +x$ $"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmmml)"+x"color(white)(mmm)"+x"$
${E/mol\cdotL}^{-1} : l l 0.35 - x m m m m m m m x m m m l l x$ $"E/mol·L"^"-1":color(white)(ll)"0.35 -"xcolor(white)(mmmmmmm)xcolor(white)(mmmll)x$

The $K_{a}$ $K_"a"$ expression is

$K_{a} = \frac{[H_{3} O^{+}] [F^{-}]}{[HF]} = 6.8 \times 10^{-4}$ $K_"a" = (["H"_3"O"^+]["F"^"-"])/(["HF"]) = 6.8 × 10^"-4"$

$K_{a} = \frac{x \times x}{0.35 - x} = \frac{x^{2}}{0.35 - x} = 6.8 \times 10^{-4}$ $K_a = (x × x)/(0.35-x) = x^2/(0.35-x) = 6.8 × 10^"-4"$

$\frac{0.35}{K_{a}} = \frac{0.35}{6.31 \times 10^{-4}} = 515 > 400$ $0.35/K_"a" = 0.35/(6.31 × 10^"-4") = 515 > 400$

∴ $x ≪ 0.35$ $x ≪ 0.35$ , and the equation becomes

$\frac{x^{2}}{0.35} = 6.8 \times 10^{-4}$ $x^2/0.35 = 6.8 × 10^"-4"$

$x^{2} = 0.35 \times 6.8 \times 10^{-4} = 2.38 \times 10^{-4}$ $x^2 = 0.35 × 6.8 × 10^"-4" = 2.38 × 10^"-4"$

$x = 1.54 \times 10^{-2}$ $x = 1.54 × 10^"-2"$

$[H^{+}] = 1.54 \times 10^{-2} l mol/L$ $["H"^+] = 1.54 × 10^"-2"color(white)(l) "mol/L"$

$pH = -log [H_{3} O^{+}] = -log (1.54 \times 10^{-2}) = 1.81$ $"pH" = "-log"["H"_3"O"^"+"] = "-log"(1.54 × 10^"-2") = 1.81$

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Question #721ab

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