• anonymous
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
  • Stacey Warren - Expert
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  • chestercat
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  • Jhannybean
\[\sf pH+pOH=14\]\[\sf pOH = 14 - pH = 14 - 9 = 5\]\[\sf pOH = 5\]\[\sf 10^{-pOH} =10^{-5}\]\[\sf [OH^-] = 10^{-pOH} =10^{-5} = 1.0 ~\times 10^{-5}\]
  • anonymous
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. \[14 = pH + pOH\]

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