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Let initial volume is 1 liter.
pH = 5
So, [H+] = 10^(-pH) = 10^(-5) = 10^-5 M
After adding water 10 folded, volume = 1*10 = 10 liters
So, new [H+] = moles/new volume = 10^-5 / 10 = 10^-6 M
So, new pH = - log [H+] = - log [10^-6] = 6
New pH = 6 ....Answer
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