Question

3.5 Exercise M -Relationship between concentrations, pH an Complete the following table for solutions at 259%C acidic, basic, neutral 2 8.32 1.00 4. 5.0x 10 1.0 x 10-12 Exercise J-Identifying acidic solutions by pH Circle the most acidic pH in each group: 2.5 1.5 10 2 12 0.9 1.5 1.3 7.4 12 7 10 5. 7.5 2.4 3.2 6. 8.5 3.8 11.2 7.6

Answer 1

Answer:
Formulae and criteria applied for calculations are,

pH = -log[H⁺] and [H⁺] = 10^-pH.

pOH = -log[HO^-] and [HO^-] = 10^-pOH.

pH + pOH = 14.00

pH < 7 : Acidic

pH = 7: Neutral

pH > 7: Basic

1) Given: pH = 8.32 (pH > 7 : Basic)

$\therefore$ pOH = 14.00 - pH = 14.00 - 8.32 = 5.68

$\therefore$ [H⁺] = 10^-8.32. = 4.8*10^-9 M

$\therefore$ [HO^-] = 10^-5.68. = 2.1*10^-6 M

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2) SOlution Neutral.

pH = pOH = 7

$\therefore$ [H⁺] = [HO^-] = 1.00*10^-7 M

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3) Given: pH = 1.00 (pH < 7 : Acidic)

$\therefore$ pOH = 14.00 - pH = 14.00 - 1.00 = 13.00

$\therefore$ [H⁺] = 10^-1. = 1.0*10^-1 M = 0.1 M

$\therefore$ [HO^-] = 10^-13. = 1.0*10^-13 M

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4) Given: [H⁺] = 5.00*10^-8 M

pH = -log[H⁺] = -log(5.00*10^-8) = 7.30 (pH > 7.00: Basic)

pOH = 14.00 - 7.30 = 6.70

[HO^-] = 10^-pOH = 10^-6.70 = 2.0*10^-7 M

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e) Given: [HO^-] = 1.0*10^-12 M

pOH = -log[HO^-] = -log(1.0*10^-12) = 12

pH = 14 - 12 = 2 (pH < 7: Acidic)

[H⁺] = 10^-pH = 10^-2 = 0.01 M

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