Answer
$[H_3O^+] = 1.5 \times 10^{- 2}M$
$[OH^-] = 6.7 \times 10^{-13}M$
Work Step by Step
1. Use the following equation to calculate the hydronium ion concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 1.82}$
$[H_3O^+] = 1.5 \times 10^{- 2}M$
** There are only two digits after the decimal point on $1.82$, therefore, the number of significant figures of the $[H_3O^+]$ must be equal to 2.
2. Write the $K_w$ expression for the water equilibrium. Then substitute the value for $[H_3O^+]$, and solve for $[OH^-]$:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$1.5 \times 10^{-2} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{1.5 \times 10^{-2}}$
$[OH^-] = 6.7 \times 10^{-13}M$
