Answer
a. $[OH^-]=1.0*10^{-7}\text{ M}$, Neutral
b. $[OH^-]=1.2*10^1\text{ M}$, Basic
c. $[OH^-]=8.2*10^{-16}\text{ M}$, Acidic
d. $[OH^-]=1.9*10^{-10}\text{ M}$, Acidic
Work Step by Step
a.
$[H^+][OH^-]=1.0*10^{-14}$
$1.0*10^{-7}[OH^-]=1.0*10^{-14}$
$[OH^-]=1.0*10^{-7}\text{ M}$
$pH = -log[H^+]=-log(1.0*10^-7)=7$
Therefore, the solution is neutral.
b.
$[H^+][OH^-]=1.0*10^{-14}$
$8.3*10^{-16}[OH^-]=1.0*10^{-14}$
$[OH^-]=1.2*10^1\text{ M}$
Because $[OH^-]>[H^+]$, the solution is basic.
c.
$[H^+][OH^-]=1.0*10^{-14}$
$12[OH^-]=1.0*10^{-14}$
$[OH^-]=8.2*10^{-16}\text{ M}$
Because $[OH^-]
