Question

what is the ph of a $1.0 \times 10^{-4} m$ koh solution?

$\bigcirc$ a. 10
$\bigcirc$ b. 3.0
$\bigcirc$ c. 11
$\bigcirc$ d. 4.0

Explanation:

Step1: Determine \([OH^-]\) from KOH dissociation

KOH is a strong base, so it dissociates completely: \(KOH
ightarrow K^+ + OH^-\). Thus, \([OH^-] = [KOH] = 1.0 \times 10^{-4}\ M\).

Step2: Calculate pOH

The formula for pOH is \(pOH = -\log[OH^-]\). Substituting \([OH^-] = 1.0 \times 10^{-4}\ M\):
\(pOH = -\log(1.0 \times 10^{-4}) = 4\).

Step3: Relate pOH to pH

We know that \(pH + pOH = 14\) (at 25°C). Solving for pH:
\(pH = 14 - pOH = 14 - 4 = 10\).

Answer:

A. 10