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Chemistry · EQUILIBRIUM
हिन्दीThe ionic product of water at \( 310\text{ K} \) is \( 2.56 \times 10^{-14}\text{ mol}^2\text{L}^{-2} \). What is the \( \text{pH} \) of a neutral aqueous solution at this temperature? (Given: \( \log 1.6 = 0.204 \))
Correct answer · Option 2
6.80
Explanation
In neutral water, \( [\text{H}^+] = \sqrt{K_w} = \sqrt{2.56 \times 10^{-14}} = 1.6 \times 10^{-7}\text{ M} \). Thus, \( \text{pH} = -\log(1.6 \times 10^{-7}) = 7 - 0.204 = 6.796 \approx 6.80 \).
