Oxidation states simplify the whole process of working out what is being oxidised and what is being reduced in redox reactions. However, for the purposes of this introduction, it would be helpful if you knew about:
<span>oxidation and reduction in terms of electron transferelectron-half-equations</span>
Answer:
The answer to your question is below
Explanation:
Data
pOH = 5.9
a) [OH⁻] = ?
pOH measures the [OH⁻]
Formula
pOH = -log[OH⁻]
-Substitution
5.9 = -log[OH⁻]
[OH⁻] = antilog (-5.9)
-Result
[OH⁻] = 1.26 x 10⁻⁶ M
b) [H₃O⁺]
pH + pOH = 14
-Solve for pH
pH = 14 - pOH
-Substitution
pH = 14 - 5.9
-Result
pH = 8.1
-Calculate [H₃O⁺]
pH = -log[H₃O⁺]
-Substitution
8.1 = -log[H₃O⁺]
[H₃O⁺] = antilog(-8.1)
-Result
[H₃O⁺] = 7.9 x 10⁻⁹ M
c) This solution is alkaline because the pH is higher than 7.
The formula is Letter A) Li2<span>O
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