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0.002(M) KOH...

0.002(M) KOH

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In 0.002 (M) KOH solution, `[OH^(-)]=0.002(M)`
`therefore pOH=-log_(10)(0.002)=2.7`
and pH=`14-2.7=11.3`
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CHHAYA PUBLICATION-EQUILIBRIUM-SOLVED NCERT EXERCISE
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  2. Calculate the pH of 0.002(M) HBr solution

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