Subjects
Science TNSB Mentoring
Class 10 Crash course
Types of chemical reactions
Ionic product of water

9. Calculation of pOH

Question:

5 m.

Answer the following question in detail:

The hydroxyl ion concentration of a solution is \(1\times10^{-9}\) \(M\). What is the \(pOH\) of the solution?

\(pOH = -log_{10}[H^+]\)
\(pOH = log_{10}[OH^-]\)
\(pOH = -log_{10}[OH^-]\)
\(pOH = log_{10}[H^+]\)

\(pOH = -log_{10}[1\times10^{-9}]\)

\(pOH = -(log_{10}\times1.0 + log_{10}\times10^{-1})\)
\(pOH = -(log_{10}\times1.0 - log_{10}\times10^{-9})\)
\(pOH = -(log_{10}\times1.0 - log_{10}\times10^{-6})\)
\(pOH = -(log_{10}\times1.0 + log_{10}\times10^{-9})\)

\(pOH = -(2 - 9\,log_{10}\,10)\)
\(pOH = (0 - 9\,log_{10}\,10)\)
\(pOH = (0 - 9\,log_{10}\,10)\)
\(pOH = -(0 - 9\,log_{10}\,10)\)

\(pOH = -(0 - 9)\)
\(pOH = (0 - 9)\)
\(pOH = -(0 - 9+2)\)
\(pOH = -(0 - 9+1)\)

\(pOH = 9\)
\(pOH = 10\)
\(pOH = - 9\)
\(pOH = 11\)

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