Question

Calculate the [H⁺] and the [OH⁻] for the following solutions: (1) pH = 3, (2) pH = 7.5, (3) pOH = 5.6

Answer 1

1. pH = 3:

`\([H^+] = 0.001\)`

`\([OH^-] = 10^(-11)\)`

2. pH = 7.5:

`\([H^+] = 3.16 * 10^(-8)\)`

`\([OH^-] = 10^(-(14-7.5))\)`

3. pOH = 5.6:

`\([OH^-] = 10^(-5.6)\)`

`\([H^+] = 10^(-(14-5.6))\)`

Explanation:

The pH and pOH of a solution are related to the concentrations of hydrogen ions
`\([H^+]\)` and hydroxide ions
`\([OH^-]\)` in the solution. The pH and pOH are related by the following equations:

`\[ \text{pH} = -\log_(10)[H^+] \]`

`\[ \text{pOH} = -\log_(10)[OH^-] \]`

To find the concentrations of
`\([H^+]\)` and
`\([OH^-]\)`, we can use the following relationships:

`\[ [H^+] = 10^{-\text{pH}} \]`

`\[ [OH^-] = 10^{-\text{pOH}} \]`

Now, let's calculate
`\([H^+]\)` and
`\([OH^-]\)` for the given solutions:

1. For pH = 3:

`\[ [H^+] = 10^(-3) = 0.001 \]`

`\[ [OH^-] = 10^(-(14-3)) = 10^(-11) \]`

2. For pH = 7.5:

\[ [H^+] = 10^{-7.5} \]

\[ [OH^-] = 10^{-(14-7.5)} \]

3. For pOH = 5.6:

`\[ [OH^-] = 10^(-5.6) \]`

`\[ [H^+] = 10^(-(14-5.6)) \]`

Please note that the concentration of hydrogen ions
`(\([H^+]\))` and hydroxide ions
`(\([OH^-]\))` always multiply to give
`\(1.0 * 10^(-14)\)` at 25 degrees Celsius for neutral water.