What are the [H+] and [OH-] for a substance with a ph of 9.2?

The answer to your question is [H⁺] = 6.31 x 10⁻¹⁰ M    [OH⁻] = 1.58 x 10⁻⁵ M      

Explanation:

Data

[H⁺] = ?

[OH⁻] = ?

pH = 9.2

Process

To find the answer to this question we must start by using the formula of pH and simplify to find the [H⁺] then calculate the pOH to find the [OH⁻]

1.- Write the formula of pH

     pH = -log [H⁺]

2.- Find the [H⁺]

    9.2 = -log[H⁺]

-Solve for [H⁺]

    [H⁺] = antilog (-9.2)

-Result

    [H⁺] = 6.31 x 10⁻¹⁰ M

3.- Calculate the pOH

     pH + pOH = 14

-Solve for pOH

     pOH = 14 - 9.2

-Result

     pOH = 4.8

4.- Calculate the [OH⁻]

-Substitution

     4.8 = -log [OH⁻]

-Solve for [OH⁻]

      [OH⁻] = antilog (-4.8)

-result

     [OH⁻] = 1.58 x 10⁻⁵ M          

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.

a is the answer to this question

explanation: