a solution of koh is 2.5 x 10^(-5) m. what is the h⁺ concentration? (hint: find poh, then ph, then h⁺.)

Explanation:

Step1: Calculate pOH

KOH is a strong - base, so $[OH^-]=2.5\times 10^{-5}\ M$. The formula for pOH is $pOH =-\log[OH^-]$. $pOH=-\log(2.5\times 10^{-5})$. Using the logarithm property $\log(ab)=\log a+\log b$, we have $pOH = - (\log2.5+\log(10^{-5}))$. Since $\log(10^{-5})=- 5$ and $\log2.5\approx0.40$, then $pOH = 5 - 0.40=4.60$.

Step2: Calculate pH

We know the relationship $pH + pOH=14$. So, $pH=14 - pOH$. $pH = 14 - 4.60 = 9.40$.

Step3: Calculate $[H^+]$

The formula for pH is $pH=-\log[H^+]$, then $[H^+]=10^{-pH}$. $[H^+]=10^{-9.40}=3.98\times 10^{-10}\ M$.

Answer:

$3.98\times 10^{-10}$