HSC question help (1 Viewer)

jimmysmith560 · Jun 28, 2022

Would the following working help?

$K_w=\left[H_3O^+\right]\left[OH^-\right]$

In a neutral solution such as water, $\left[H_3O^+\right]=\left[OH^-\right]$

$\rightarrow \:5.5\times 10^{-14}=\left[H_3O^+\right]^2$

$\left[H_3O^+\right]=2.3\times 10^{-7}\text{mol}\:\text{L}^{-1}$

$pH=-\log _{10}\left(2.3\times 10^{-7}\right)$

$=6.63,\:\text{i.e (B).}$

astraaaaaah · Jun 28, 2022

thanks so much

HSC question help (1 Viewer)

astraaaaaah

New Member

jimmysmith560

Le Phénix Trilingue

astraaaaaah

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)