Question

Determine the [OH−]

, pH, and pOH of a solution with a [H+]
of 4.7×10−5 M
at 25 °C.

Answer 1

• 
  `\([OH^-}] = 2.04 * 10^(-5) \, \text{M}\)`
• 
  `\( \text{pH} \ = 9.67\)`

• 
  `\( \text{pOH} \ = 4.33\)`

To determine the [OH⁻], pH, and pOH of a solution with a given [H⁺], you can use the following relationships:

• OH⁻ and pOH Relationship:

pOH = -log[OH⁻]

• [H⁺] and pH Relationship:

pH = -log[H⁺]

• pH and pOH Relationship:

pH} + pOH = 14

Given [H⁺] =
`4.7 * 10^(-5)\)` M, we can calculate the corresponding values:

`\[ \text{pOH} = -\log(4.7 * 10^(-5)) \]`

`\[ \text{pOH} \approx 4.33 \]`

`\[ \text{pOH} + \text{pH} = 14 \]`

`\[ \text{pH} = 14 - \text{pOH} \]`

`\[ \text{pH} \approx 14 - 4.33 \]`

`\[ \text{pH} \approx 9.67 \]`

[OH⁻]:

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

4.33 = -log[OH⁻]

[OH⁻] =
`10^(-4.33)`

`\[ [\{OH^-}] \approx 2.04 * 10^(-5) \, \text{M} \]`

So, for the given solution:

• 
  `\([OH^-}] = 2.04 * 10^(-5) \, \text{M}\)`
• 
  `\( \text{pH} \ = 9.67\)`

• 
  `\( \text{pOH} \ = 4.33\)`

Answer 2

Step-by-step explanation:

[H+] = 4.7 x 10⁻⁵ M

[H+][OH⁻] = 10⁻¹⁴

[OH⁻] =
`(10^(-14))/(4.7* 10^(-5))`

= .2127 x 10⁻⁹ M

pH = - log[H+]

= - log [ 4.7 x 10⁻⁵]

= - log4.7 + 5

= 4.328

pH + pOH = 14

pOH = 14 - pH

= 14 - 4.328

= 9.672