A solution has a pOH of 10.65. What is the [OH-] of the solution? [OH-] = [?] x 10²] M Coefficient (green) Exponent (yellow) Enter

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### Determining Hydroxide Ion Concentration from pOH

#### Problem Statement:
A solution has a pOH of 10.65. What is the \([OH^{-}]\) (hydroxide ion concentration) of the solution?

\[ [OH^{-}] = [ ? ] \times 10^{[ ? ]} \, \text{M} \]

#### Interactive Input:
Below the equation, there are two input fields:
- The first field is labeled "Coefficient (green)"
- The second field is labeled "Exponent (yellow)"

There is also an "Enter" button to submit the values.

#### Explanation:

1. **Understanding pOH and \([OH^{-}]\):**
   - pOH is a measure of the alkalinity (basicity) of a solution.
   - The relationship between pOH and \([OH^{-}]\) is given by the equation: \( pOH = -\log[OH^{-}] \).

2. **Calculating \([OH^{-}]\):**
   - To find \([OH^{-}]\), we rearrange the formula to: \([OH^{-}] = 10^{-pOH} \).
   - Given that pOH = 10.65, calculate \([OH^{-}]\) as follows:
     \[ [OH^{-}] = 10^{-10.65} \]

3. **Converting to the Required Format:**
   - The output format suggests expressing \([OH^{-}]\) in scientific notation: \([OH^{-}] = a \times 10^{-b} \, \text{M} \).
   - Compute the values where \( a \) is the coefficient and \( -b \) is the exponent.

#### Interactive Step:

To solve the problem, input the coefficient and exponent in their respective fields and click "Enter" to verify the solution.

This educational module helps reinforce understanding of logarithmic relationships and scientific notation, essential for grasping concepts in chemistry.
Transcribed Image Text:### Determining Hydroxide Ion Concentration from pOH #### Problem Statement: A solution has a pOH of 10.65. What is the \([OH^{-}]\) (hydroxide ion concentration) of the solution? \[ [OH^{-}] = [ ? ] \times 10^{[ ? ]} \, \text{M} \] #### Interactive Input: Below the equation, there are two input fields: - The first field is labeled "Coefficient (green)" - The second field is labeled "Exponent (yellow)" There is also an "Enter" button to submit the values. #### Explanation: 1. **Understanding pOH and \([OH^{-}]\):** - pOH is a measure of the alkalinity (basicity) of a solution. - The relationship between pOH and \([OH^{-}]\) is given by the equation: \( pOH = -\log[OH^{-}] \). 2. **Calculating \([OH^{-}]\):** - To find \([OH^{-}]\), we rearrange the formula to: \([OH^{-}] = 10^{-pOH} \). - Given that pOH = 10.65, calculate \([OH^{-}]\) as follows: \[ [OH^{-}] = 10^{-10.65} \] 3. **Converting to the Required Format:** - The output format suggests expressing \([OH^{-}]\) in scientific notation: \([OH^{-}] = a \times 10^{-b} \, \text{M} \). - Compute the values where \( a \) is the coefficient and \( -b \) is the exponent. #### Interactive Step: To solve the problem, input the coefficient and exponent in their respective fields and click "Enter" to verify the solution. This educational module helps reinforce understanding of logarithmic relationships and scientific notation, essential for grasping concepts in chemistry.
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