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10 cards from this deck
Hydroxide ion concentration
pOH=−log10[OH−]\text{pOH} = -\log_{10}[\text{OH}^-]pOH=−log10[OH−]
[OH−]=10−pOH[\text{OH}^-] = 10^{-\text{pOH}}[OH−]=10−pOH
pH+pOH=14\text{pH} + \text{pOH} = 14pH+pOH=14
1.0×10−141.0 \times 10^{-14}1.0×10−14
1:1 or [NaOH]=[OH−][\text{NaOH}] = [\text{OH}^-][NaOH]=[OH−]
1:2 or [Ba(OH)2]=12[OH−][\text{Ba(OH)}_2] = \frac{1}{2}[\text{OH}^-][Ba(OH)2]=21[OH−]
1 unit
2 units
25°C or 298 K
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