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The ionic product of water at 310 K is 2.7 × . What is the pH of neutral water at this temperature? |
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Answer» roduct, KW = [H+][OH-] Let's ASSUME that [H+] = x Now, as we know that [H+] = [OH-] We get, Kw = x2 = 2.7 × 10–14 THEREFORE, we get, x = 1.64 × 10–7 Also, pH = -log[H+] =-log(1.64 × 10–7) = 6.78 |
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