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10 questions from this quiz
pOH=−log10[OH−]\text{pOH} = -\log_{10}[\text{OH}^-]pOH=−log10[OH−]
3.82
1.6×10−61.6 \times 10^{-6}1.6×10−6 mol L−1^{-1}−1
[NaOH]=[OH−][\text{NaOH}] = [\text{OH}^-][NaOH]=[OH−]
[Ba(OH)2]=12[OH−][\text{Ba(OH)}_2] = \frac{1}{2}[\text{OH}^-][Ba(OH)2]=21[OH−]
pH+pOH=14\text{pH} + \text{pOH} = 14pH+pOH=14
10.1
5
1.0×10−141.0 \times 10^{-14}1.0×10−14
pOH decreases
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