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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1. 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}$

2. 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. $14 = pH + pOH$