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Question

Chemistry Question on Equilibrium

At 90∘C,90^\circ C, pure water has [H3O+][H_3 O^+] as 10−6 mol L−1 10^{ - 6 } \, mol \, L^{ - 1}. What is the value of Kw at 90∘K_w \, at \, 90^\circ C ?

A

10−610^{ - 6 }

B

10−1210^{ - 12 }

C

10−1410^{ - 14 }

D

10−810^{ - 8 }

Answer

10−1210^{ - 12 }

Explanation

Solution

Kw=[H3O+] [OH−]=10−6×10−6=10−12K_w = [ H_3 O^+ ] \, [ OH^- ] = 10^{ - 6 } \times 10^{ - 6 } = 10^{ - 12 }