Question

find the ph of a solution containing 0.45 m of the strong base naoh.
0.35
13.65
15.13
-1.13

Explanation:

Step1: Determine $[OH^-]$ concentration

Since $NaOH$ is a strong - base and dissociates completely as $NaOH
ightarrow Na^+ + OH^-$, the concentration of $OH^-$ ions is equal to the concentration of $NaOH$. So, $[OH^-]=0.45\ M$.

Step2: Calculate pOH

The formula for pOH is $pOH =-\log[OH^-]$. Substituting $[OH^-]=0.45\ M$ into the formula, we get $pOH=-\log(0.45)\approx0.35$.

Step3: Calculate pH

We know the relationship $pH + pOH=14$. Rearranging for pH gives $pH = 14 - pOH$. Substituting $pOH\approx0.35$ into the formula, we get $pH=14 - 0.35 = 13.65$.

Answer:

13.65