pH of solution at 25 degree Celsius which has twice as many hydroxide and as in pure water 25 degree Celsius will be
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Answer:
Ka × Kb = 10^-14
Since pure water has concentration of 10^-7 hydroxide ions, this solution will have 2 × 10^-7 hydrixide ions.
Therefore as per the above mentioned equation, the value of Ka would be 0.5 × 10^-7 and its Ph would be -log[0.5×10^-7] = -log(0.5) + 7 = 0.3010 + 7 = 7.3010 .
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