Answer
a. $[H^+]$ and $[OH^-] = 1.709\times 10^{- 7}$
b. It is equal to $6.767$.
c. $pH$ = $12.530$
Work Step by Step
a.
- Pure water solution : $[H^+] = [OH^-]$
- Create a unknown called 'x', that has the value of $[H^+]\ and\ [OH^-]$
- Write the $K_w$ expression, and calculate the x value:
$[H^+] * [OH^-] = K_w = 2.92\times 10^{- 14}$
$x * x = 2.92\times 10^{- 14}$
$x^2 = 2.92\times 10^{- 14}$
$x = \sqrt { 2.92\times 10^{- 14}}$
$x = 1.71\times 10^{- 7}$
Therefore: $[H^+]\ and\ [OH^-] = 1.71\times 10^{- 7}M$
b.
$pH = -log[H^+]$
$pH = -log( 1.71 \times 10^{- 7})$
$pH = 6.767$
c.
$[OH^-] * [H^+] = Kw = 2.92 \times 10^{-14}$
$ 0.1 * [H^+] = 2.92 \times 10^{-14}$
$[H^+] = \frac{2.92 \times 10^{-14}}{0.1}$
$[H^+] = 2.92 \times 10^{-13}M$
$pH = -log[H^+]$
$pH = -log( 2.92 \times 10^{- 13})$
$pH = 12.53$
