1 c.c of 0.1N HCL is added to 1L solution of NACL. the pH of resulting solutionwill be??

Volume of HCl = 1 cm3‚Äč or 0.001 L
Normality of HCl = 0.1 N
Number of gram-equivalents of HCl = Normality × Volume
= 0.1 × 0.001
= 0.0001
Volume of solution = Volume of NaCl + Volume of HCl
= 1 + 0.001
= 1.001 L
As NaCl is salt it will not contribute towards the pH of the solution.
Normality of HCl in resulting solution = 0.0001 / 1.001
= 0.0001 N
pH = -log[H+]
= - log(0.0001)
= 4 

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