⦁ Find the concentration of H+, OH-, PH and POH of 0.03 M of magnesium hydroxide which ionizes to the extent of only 1 /3 in aqueous solution.

Answers

Answer 1

Answer:

[tex]pH=12.3\\\\pOH=1.7\\[/tex]

[tex][H^+]=5x10^{-13}M[/tex]

[tex][OH^-]=0.02M[/tex]

Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

[tex]Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)[/tex]

Thus, since the ionization occurs at an extent of 1/3, we can set  up the following relationship:

[tex]\frac{1}{3} =\frac{x}{[Mg(OH)_2]}[/tex]

Thus, x for this problem is:

[tex]x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M[/tex]

Now, according to an ICE table, we have that:

[tex][OH^-]=2x=2*0.01M=0.02M[/tex]

Therefore, we can calculate the H^+, pH and pOH now:

[tex][H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M[/tex]

[tex]pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7[/tex]

Best regards!


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Answer:

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For the reaction represented by the equation:

Cl2 + 2KBr → Br2 + 2KCl

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Answers

Answer:

223 g KCl

General Formulas and Concepts:

Atomic Structure

Reading a Periodic TableMoles

Stoichiometry

Using Dimensional AnalysisAnalyzing reactions RxN

Explanation:

Step 1: Define

[RxN - Balanced] Cl₂ + 2KBr → Br₂ + 2KCl

[Given] 356 gg KBr

[Solve] g KCl

Step 2: Identify Conversions

[RxN] 2 mol KBr → 2 mol KCl

[PT] Molar Mass of K - 39.10 g/mol

[PT] Molar Mass of Br - 79.90 g/mol

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of KBr - 39.10 + 79.90 = 119 g/mol

Molar Mass of KCl - 39.10 + 35.45 = 74.55 g/mol

Step 3: Stoichiometry

[DA] Set up conversion:                                                                                   [tex]\displaystyle 356 \ g \ KBr(\frac{1 \ mol \ KBr}{119 \ g \ KBr})(\frac{2 \ mol \ KCl}{2 \ mol \ KBr})(\frac{74.55 \ g \ KCl}{1 \ mol \ KCl})[/tex][DA] Divide/Multiply [Cancel out units]:                                                           [tex]\displaystyle 223.024 \ g \ KCl[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

223.024 g KCl ≈ 223 g KCl

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Answers

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Givens

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Solution

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4

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