Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryIonic Equilibria3 Marks
Question
Derive the relation pH + pOH = 14.
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Answer
The ionic product of water, Kw is given by, $KW =\left[ H _3 O ^{+}\right] \times\left[ OH ^{-}\right]$ At $298 K _{,} K _{ w }=1 \times 10^{-14}$ $\therefore pK _{ w }=-\log _{10} K _{ w }=\log _{10} 1 \times 10^{-14}=14$ $\because\left[ H _3 O ^{+}\right] \times\left[ OH ^{-}\right]=1 \times 10^{-14}$ Taking logarithm to base 10 of both sides, $\log _{10}\left[ H _3 O ^{+}\right]+\log _{10}\left[ OH ^{-}\right]=\log _{10} 1 \times 10^{-14}$ Multiplying both the sides by -1 , $-\log _{10}\left[ H _3 O ^{+}\right]-\log _{10}\left[ OH ^{-}\right]=-\log _{10} 1 \times 10^{-14}$ $\because pH =-\log _{10}\left[ H _3 O ^{+}\right] ; pOH =-\log _{10}\left[ OH ^{-}\right] ;$ pKw $=-\log _{10} K _{ w }$ $\therefore pH + pOH = pK _{ w }$ OR $ pH + pOH =14$
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