Kw of water at 373k is 1*10-12. what will be ph of h20 at 373 k ? is water acidic basic or neutral at this temperature
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Ionic product, 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
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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