Example 7

This seems to be a pretty easy problem but I cannot understand why we do not consider the contribution of $\ce{H+}$ ions by water (water dissociates into $\ce{H+}$ and $\ce{OH-}$) and only that of $\ce{HCl}$ when we find the total $\ce{H+}$ ion concentration, in the same way both $\ce{NaCl}$ and $\ce{HCl}$ contribute towards the concentration of Chlorine.

Is the solution wrong or is my reasoning incorrect?

  • $\begingroup$ It'd be nice if you transcribed the image into text since images are not searchable. $\endgroup$ – Buck Thorn Nov 20 '19 at 8:54
  • $\begingroup$ What kind of old textbook ist that? Please add a citation. The usage of normality is deprecated for years, decades even. $\endgroup$ – Martin - マーチン Nov 20 '19 at 9:00
  • $\begingroup$ @Martin A Book of physical chemistry by OP Tandon. Its Indian. $\endgroup$ – John Tony Nov 20 '19 at 9:02

The contribution by water would be of the order $\pu{10^{-7} M}$ at Room Temperature which is negligible in comparison to that provided by $\ce{HCl}$ .

In the second case the contributions by both $\ce{NaCl}$ and $\ce{HCl}$ towards Chlorine Ions are comparable hence both need to be included.

However nevertheless one may at times need to account for contributions from Water say for instance one needs to calculate the pH of $\pu{10^{-8} M}$ $\ce{HCl}$ as here it wouldn't be negligible but instead comparable.

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  • $\begingroup$ oops. You've already answered. I'll delete mine. $\endgroup$ – William R. Ebenezer Nov 20 '19 at 8:51
  • $\begingroup$ But if we consider that Water has 15 milli equivalents, its concentration will be 15/100 = 0.15M . Won't its dissociation into H+ and OH- lead to the formation of H+ with a similar concentration? $\endgroup$ – John Tony Nov 20 '19 at 8:51
  • $\begingroup$ Or is its degree of dissociation just very low? $\endgroup$ – John Tony Nov 20 '19 at 8:52
  • $\begingroup$ @JohnTony Yes, quite low. $\endgroup$ – William R. Ebenezer Nov 20 '19 at 8:53
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    $\begingroup$ @William R. Ebenezer Nice to see an IITian here! Hope to be one in the future. $\endgroup$ – John Tony Nov 20 '19 at 8:58

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