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anonymous
 one year ago
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
anonymous
 one year ago
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

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Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4\[\sf pH+pOH=14\]\[\sf pOH = 14  pH = 14  9 = 5\]\[\sf pOH = 5\]\[\sf 10^{pOH} =10^{5}\]\[\sf [OH^] = 10^{pOH} =10^{5} = 1.0 ~\times 10^{5}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We 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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