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# Ionic Product of Water at 310 K is 2.7 × 10–14. What is the Ph of Neutral Water at this Temperature? - CBSE (Science) Class 11 - Chemistry

ConceptIonization of Acids and Bases The Ionization Constant of Water and Its Ionic Product

#### Question

Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?

#### Solution

Ionic product,

K_w = [H^+] [OH^-]

Let [H^+] = x

Since [H^+] = [OH^(-)], K_w = x^2

=> K_w "at 310K"  is 2.7  xx 10^(-14)

:. 2.7 xx 10^(-14) = x^2

=> x = 1.64 xx 10^(-7)

=> [H^+] = 1.64 xx 10^(-7)

=> pH = -log[H^+]

= -log [1.64 xx 10^(-7)]

= 6.78

Hence, the pH of neutral water is 6.78

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#### APPEARS IN

Solution Ionic Product of Water at 310 K is 2.7 × 10–14. What is the Ph of Neutral Water at this Temperature? Concept: Ionization of Acids and Bases - The Ionization Constant of Water and Its Ionic Product.
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