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If a solution of Sulfuric acid H2SO4 has a concentration of 2.1 x 10-4 M or 2.1E-4 M, what is the concentration of its hydroxide ion?

In a given solution, the concentration of hydrogen ions and hydroxide ions are related to each other by the following expression:


pH + pOH = 14


where, pH = - log[H+] and pOH = - log [OH-]


In this case, the concentration of sulfuric acid is given as 2.1 x 10^-4 M. Since each molecule of sulfuric acid (H2SO4) contains 2 atoms of hydrogen, the concentration of hydrogen ions in this solution is twice that...

In a given solution, the concentration of hydrogen ions and hydroxide ions are related to each other by the following expression:


pH + pOH = 14


where, pH = - log[H+] and pOH = - log [OH-]


In this case, the concentration of sulfuric acid is given as 2.1 x 10^-4 M. Since each molecule of sulfuric acid (H2SO4) contains 2 atoms of hydrogen, the concentration of hydrogen ions in this solution is twice that of sulfuric acid. That is,


Concentration of H+ ions = 2 x 2.1 x 10^-4 


= 4.2 x 10^-4 M


This means, pH = -log (4.2 x 10^-4) = 3.38


Since pH + pOH = 14


pOH = 14 - 3.38 = 10.62


This means, 10.62 = -log [OH-]


Solving the equation, we get [OH-] = 2.4 x 10^-11 M.


Hope this helps.

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