Question

1ml of 0.01 n hcl is added to 999ml solution of 0.1 n na2so4 the ph of the resulting solution will

Answer 1

Answer:

Volume of  HCl  =0.001L,    N  of HCl  =0.1N

Volume of  NaCl  =0.999L

Volume of  solution  =Volume of   HCl  +Volume of   NaCl  =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