Question

# The Kw of $${ 2H }_{ 2 }{ O }\quad \rightleftarrows { H }_{ 3 }{ O }^{ + }+O{ H }^{ - }$$ changes from $${ 10 }^{ -14 }$$ at $${ 25 }^{ \circ }C$$ to $$9.62 \times { 10 }^{ -14 }$$ at $${ 60 }^{ \circ }$$. What is the pH of water at $${ 60 }^{ \circ }C$$?

Solution

## $${ K }_{ w }=\left[ { H }^{ + } \right] \quad \left[ { OH }^{ - } \right] \\ \quad \quad \quad \left[ { H }^{ + } \right] \quad \left[ { OH }^{ - } \right] \quad \\ \quad \quad [{ H }^{ + }{ ] }^{ 2 }=9.62\times { 10 }^{ -14 }\\ { H }^{ + }=\sqrt { 9.62\times { 10 }^{ -14 } } =\quad 3.1\times { 10 }^{ -7 }\\ PH=\quad -log(3.1\times { 10 }^{ -7 })\\ \quad \quad \quad =\quad -0.5+7\\ PH=\quad 6.5\\ solution\quad would\quad be\quad little\quad acidic.$$Chemistry

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