Question

what is the ph of a 1.5e - 4 m csoh solution?

Explanation:

Step1: Determine $[OH^-]$

Since $CsOH$ is a strong base and dissociates completely as $CsOH = Cs^+ + OH^-$, so $[OH^-]=1.5\times10^{-4}\ M$

Step2: Calculate $pOH$

Use the formula $pOH=-\log[OH^-]$.
$$pOH =-\log(1.5\times 10^{-4})=-\log1.5+4\approx - 0.176 + 4=3.824$$

Step3: Calculate $pH$

Use the relationship $pH + pOH=14$.
$$pH=14 - pOH=14 - 3.824 = 10.176\approx10.18$$

Answer:

$10.18$