## Chemistry: The Central Science (13th Edition)

$[OH^-]=0.022 M$ $pH=12.34$
$[OH^-]=x$ $K_b=6.4\times10^{-4}=\frac{x\times x}{0.075-x}$ We can assume that $x\lt\lt0.075$ Therefore, $x=\sqrt {6.4\times 10^-4\times 0.075}$ $x=[OH^-]=0.022 M$ $pOH=1.66$ $pH=14-pOH=12.34$