At 90∘C, pure water has [H+]=10−6 M.If 100 mL of 0.2 M HCl is added to 200 mL of 0.1 M KOH at 90∘C then pH of the resulting solution will be :

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At {90}^{{}}C, pure water has {[{H}^{{+}}]}={10}^{{-{6}}} M.If

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At $90^{\circ}$ C, pure water has $[H^{+}] = 10^{- 6}$ M.If 100 mL of $0.2$ M HCl is added to 200 mL of $0.1$ M KOH at $90^{\circ}$ C then pH of the resulting solution will be :

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90^{\circ}

C, pure water has

[H^{+}] = 10^{- 6}

M.If 100 mL of

0.2

M HCl is added to 200 mL of

0.1

M KOH at

90^{\circ}

C then pH of the resulting solution will be :

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Question Text	At $90^{\circ}$ C, pure water has $[H^{+}] = 10^{- 6}$ M.If 100 mL of $0.2$ M HCl is added to 200 mL of $0.1$ M KOH at $90^{\circ}$ C then pH of the resulting solution will be :
Answer Type	Text solution:1
Upvotes	153