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
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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?
`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)]`
Hence, the pH of neutral water is 6.78
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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.