HA is a weak acid. A 0.21M solution of HA has a pH of 2.68. What is the Ka value for HA?

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HA is a weak acid. A 0.21M solution of HA has a pH of 2.68. What is the Ka value for HA?

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Answer 1 · 2018-06-18T01:16:15

$K_{a} = 2.1 \times 10^{-5}$ $K_text(a) = 2.1 × 10^"-5"$

Explanation:

$pH = 2.68$ $"pH = 2.68"$

$[H_{3} O^{+}] = 10^{-pH} l mol/L = 10^{-2.68} l mol/L = 0.002 09 mol/L$ $["H"_3"O"^"+"] = 10^"-pH" color(white)(l)"mol/L" = 10^"-2.68"color(white)(l)"mol/L" = "0.002 09 mol/L"$

We can use an ICE table to help us calculate the value of $K_{a}$ $K_text(a)$ .

$m m m m m m l l HA + H_{2} O ⇌ H_{3} O^{+} + l A^{-}$ $color(white)(mmmmmmll)"HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + color(white)(l)"A"^"-"$
${I/mol\cdotL}^{-1} : m l l 0.21 m m m m m m 0 m m m l 0$ $"I/mol·L"^"-1": color(white)(mll)0.21 color(white)(mmmmmm)0color(white)(mmml)0$
${C/mol\cdotL}^{-1} : m l - x m m m m m m l + x m m l + x$ $"C/mol·L"^"-1": color(white)(ml)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mml)"+"x$
${E/mol\cdotL}^{-1} : m 0.21- x m m m m m l x m m m l x$ $"E/mol·L"^"-1": color(white)(m)"0.21-"xcolor(white)(mmmmml)xcolor(white)(mmml)x$

At equilibrium,

$[H_{3} O^{+}] = 0.002 09 mol/L$ $["H"_3"O"^"+"] = "0.002 09 mol/L"$

∴ $x = 0.002 09$ $x = "0.002 09"$

$K_{a} = \frac{[H_{3} O^{+}] [A^{-}]}{[HA]} = \frac{x^{2}}{0.21 - x} = \frac{{0.002 09}^{2}}{0.21-0.002 09} = \frac{4.16 \times 10^{-6}}{0.208}$ $K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = x^2/(0.21-x) = "0.002 09"^2/("0.21-0.002 09") = (4.16 × 10^"-6")/0.208$

$= 2.1 \times 10^{-5}$ $= 2.1 × 10^"-5"$

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HA is a weak acid. A 0.21M solution of HA has a pH of 2.68. What is the Ka value for HA?

1 Answer

Explanation:

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