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At 90 \degree C pure water has [ H_3O^+]   as 10^{-6} mol/L  what is the value of k_w  at 90 \degree ? 

Option: 1

10^{-6}


Option: 2

10^{-12}


Option: 3

10^{-14}


Option: 4

10^{-10}


Answers (1)

best_answer

Given,

[H_3O^+] = 10^{-6} mol L^{-1}

\therefore [^-OH] = 10^{-6} molL^{-1}

K_{w}=[H^{+}]\:[\bar{O}H] = 10 ^{-6} \times 10 ^{-6} = 10^{-12} 

Therefore, Option(2) is correct

Posted by

Gaurav

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