Question

A solution with a pH of 9 has a [OH-] concentration of




A. 1.0 × 10–14 mol/L.


B. 1.0 × 10–9 mol/L.


C. 1.0 × 10–5 mol/L.


D. 1.0 × 10–7 mol/L.
hELP

Answer 1

\[\sf pH+pOH=14\]\[\sf pOH = 14 - pH = 14 - 9 = 5\]\[\sf pOH = 5\]\[\sf 10^{-pOH} =10^{-5}\]\[\sf [OH^-] = 10^{-pOH} =10^{-5} = 1.0 ~\times 10^{-5}\]

Answer 2

We could go back a little further before @Jhannybean 's post and say that we found the equilibrium constant of water to be \[K_w = 10^{-14}\] which is a super tiny number, that shows that not much water dissociates into \(H^+\) and \(OH^-\). Then we can write out the equilibrium equation:

\[K_w= [H^+][OH^-]\]
\[10^{-14}= [H^+][OH^-]\]
Now take the negative log base 10 of both sides:

\[-\log_{10} 10^{-14} = -\log_{10} ([H^+][OH^-])\]
Then use some log properties to take the -14 exponent down and separate the two terms on the right:
\[14\log_{10} 10 = -\log_{10} [H^+] -\log_{10} [OH^-]\]

Then this simplifies to the start of her answer, in case you wanted to go further back, just for fun.
\[14 = pH + pOH\]