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Gelneren [198K]
3 years ago
12

What is the approximate pH of a .06 M solution of CH3COOH given that Ka= 1.78*10-5

Chemistry
1 answer:
Step2247 [10]3 years ago
4 0
CH₃COOH ⇔ CH₃COO⁻ + H⁺

[CH₃COO⁻] = [H⁺] = x

K=\frac{|CH_{3}COO^{-}|*|H^{+}|}{|CH_{3}COOH|}=\frac{x^{2}}{|CH_{3}COOH|}\\\\
1,78*10^{-5}=\frac{x^{2}}{0,06} \ \ |*0,06\\\\
0,1068*10^{-5}=x^{2}\\\\
x_{1}\approx0,001 \ \land \ \ x_{2}=\approx-0,001\\\\
pH=-log|H^{+}|=-log0,001=3
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Using Charles law

\\ \sf\longmapsto V_1T_2=V_2T_1

\\ \sf\longmapsto V_2=V_1T_2\div T_1

\\ \sf\longmapsto V_2=\dfrac{49.8(356)}{291}

\\ \sf\longmapsto V_2=\dfrac{17728.8}{291}

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Question 12 of 25
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