Given that the dissociation constant for H2O is Kw-1- 10-14 mol2 L-2- what is the pH of 0-001 molar KOH solution-

The correct option is A 11 0.001 M KOH solution [OH^ ]=0.001 M=1× 10^ 3 M [H^+]× [OH^ ]=1× 10^ 14 [H^+]=1× 10^ 14[OH^ ] [H^+]=1× ...

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pH the Power of H

Given that th...

Question

Given that the dissociation constant for $H_{2} O$ is $K_{w} = 1 \times 10^{- 14} m o l^{2} L^{- 2}$ , what is the $p H$ of $0.001 m o l a r K O H$ solution?

10^{- 11}

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10^{- 3}

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3

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11

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C O_{3}^{2 -}

K_{a 1} f o r H_{2} C O_{3} = 4.45 \times 10^{- 7}

K_{a 2} f o r H_{2} C O_{3} = 4.69 \times 10^{- 11}

10^{- 4.18} = 6.6 \times 10^{- 5}

p H

25^{\circ}

F e (N O_{3})_{3}

1.0 \times 10^{- 3}

[F e (H_{2} O)_{6}]^{3 +} + H_{2} O (l) ⇌ H_{3} O^{+} (a q .) + [F e (H_{2} O)_{5} (O H)]^{2 +}

p H

0.1 M

N a H {C O}_{3}

K_{1} = 4.5 \times 10^{- 7}, K_{2} = 4.5 \times 10^{- 11}

p H

N i (O H)_{2}

10^{- 3} M

N i^{2 +}

[K_{s p}

(N i (O H)_{2} = 10^{- 11}]

N a C N

\frac{N}{80}

H C N

1.3 \times 10^{- 9}

K_{w} = 1.0 \times 10^{- 14}

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