Answer
$[H_3O^+] = 3 \times 10^{- 7}M$
$[OH^-] = 3 \times 10^{-8}M$
Work Step by Step
1. Use the following equation to calculate the hydronium ion concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 6.5}$
$[H_3O^+] = 3 \times 10^{- 7}M$
** There is only one digit after the decimal point on $6.5$, therefore, the number of significant figures of the $[H_3O^+]$ must be equal to 1.
2. Write the $K_w$ expression for the water equilibrium. Then substitute the value for $[H_3O^+]$, and solve for $[OH^-]$:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$3 \times 10^{-7} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{3 \times 10^{-7}}$
$[OH^-] = 3 \times 10^{-8}M$
