Question

What is the pH, H3O+, and -OH of an aqueous 7.5×10^-2 M RbOH solution at 25c? pH = -logH3O+ and Kw = 1.0 × 10^-14 = H3O+ [-OH] at 25c.

Answer 1

RbOH is a strong base, so it dissociates in water completely. As 1 mole of RbOH produces, 1 mol of hydroxide ions.

Thus, $[OH^-]=[RbOH]=7.5*10^{-2}M$

As this concentration is much greater than the concentration of the concentration of hydroxide ions contributed by water.

$pOH=-log([OH^-])=2-log(7.5)=2-0.875=1.125$

As $pH+pOH=14 \implies pH=14-1.125=12.875$

And thus, $[H^+]=10^{-12.875}=0.1333*10^-12=1.333*10^{-13}M$