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in Equilibrium by (15 points)
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When 900 ml of water is added in 100 ml of 0.01N HCl solution then pH of resultant solution will be:

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1 Answer

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Given that 

Volume of HCl = 100 ml = 0.1 L

Normality of HCl = 0.1 N

Now we can find

Gram-equivalent of HCl = Normality × Volume

= 0.1 × 0.1 = 0.01

So,

Volume of solution = Volume of H2​O + Volume of HCl

= 900 + 100 = 1000 mL

= 100 mL= 1 L

Normality of  HCl in resulting solution = 0.01/1 ​= 0.01 N

pH = −log[H+]

= −log(0.01)

= 2

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