1)- What is the pH of10^{-8} (N) HCl?

The pH of 10^{-8} N HCl can be calculated as follow:

[H+] = 10^{-8} N

As the solution is so dilute that water will also dissociate to produce H^{+ }ions

H_{2}O → H^{+} + OH^{-}

Now,

The Concentration of H+ released from water is 10^{-7} as the pH of water is equal to 7.

Hence the total concentration of H+ = 10^{-7} + 10^{-8}

[H^{+}] = 10^{-7} + 0.1 × 10^{-7 }

= 1.1 × 10^{-7 }

pH = -log [H^{+}]

Hence pH = 6.95

Hence the pH of 10^{-8} N solution of HCl = 6.95.

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