At 90^° C, pure water has \lbrack H2O'\rbrack = 10-6 moL L-1. What is the value of K, at 90^° C?(1) 10-6(2) 1012(3) 10-144)10-1216. The degree of dissociation of 0.1 M CH.COOH will be equal to(Given: K(CH2COOH) = 1x10)(1) 10-5(2) 10-4(3) 10-3(410-217ther havingpH = 4 are mixed. The pH of thefo

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At $90^{\circ}$ C pure water has [ ${H_{3} O}^{+}$ ] = $10^{- 6}$ mol ${l i t r e}^{- 1}$ . The value of $K_{W}$ at $90^{\circ}$ C is

90^{o} C

[H_{3} O^{+}] = 10^{- 6}

m o l / l i t r e

K_{w}

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90^{o} C

[H_{3} O^{\oplus}] = 10^{- 6.7} m o l L^{- 1}

K_{w}

90^{o} C

90^{\circ} C

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

K_{w}

90^{\circ} C

90^{\circ} C,

[H_{3} O^{+}] = 10^{- 6} m o l L^{- 1} .

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90^{\circ} C ?

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