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1ml of 0.1NHCl is added to 999ml solution of NaCl. The pH of the resulting solution will be:

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VHCl ​ =0.001L,  NHCl ​ =0.1N 

VNaCl ​ =0.999L 

Vsolution ​ =VHCl ​ +VNaCl ​ =1 

∴ Normality of HCl in a resulting solution=0.1×0.001/1 ​ =0.0001N=[H+

Addition of NaCl does not effect the pH of solution as it is a salt & is neutral. 

So, pH=−log[H+] = −log[0.0001] = 4

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