Questions: Question 4 pH + pOH always equals log(Kw) 14.0 1.00 × 10^-14 pKw

Transcript text: Question 4 $\mathrm{pH}+\mathrm{pOH}$ always equals $\log \left(K_{w}\right)$ 14.0 $1.00 \times 10^{-14}$ $\mathrm{pK}_{\mathrm{w}}$

Solution

Solution Steps

Step 1: Understanding the Relationship Between pH and pOH

The relationship between pH and pOH in aqueous solutions is given by the equation:

\[ \text{pH} + \text{pOH} = 14.0 \]

This is derived from the ion product of water at 25°C, where $K_w = 1.00 \times 10^{-14}$. The pK$_w$ is the negative logarithm of $K_w$, which is 14.0.

Step 2: Analyzing the Options

$\log(K_w)$ is not equal to $\text{pH} + \text{pOH}$.
14.0 is the correct value for $\text{pH} + \text{pOH}$ at 25°C.
$1.00 \times 10^{-14}$ is the value of $K_w$, not $\text{pH} + \text{pOH}$.
$\text{pK}_w$ is equal to 14.0, but the direct answer to $\text{pH} + \text{pOH}$ is 14.0.

Final Answer

$\boxed{14.0}$

Was this solution helpful?

Unhelpful

Helpful

Questions asked by other users

< Previous Next >

Thank you for your feedback.

Copied!