A #0.0729*g# mass of #HCl(g)# was dissolved in a #50*mL# volume of water. What is #pH# of this solution?

1 Answer
Oct 27, 2016

Answer:

#pH=-log_10[H_3O^+]=??#

#pH<3#

Explanation:

#HCl(g) + H_2O(l) rarr H_3O^+ + Cl^-#

This reaction is essentially quantitative, and the solution will be stoichiometric in #H_3O^+#.

#"Moles of HCl"# #=# #(0.0729*g)/(36.46*g*mol^-1)#

And thus #[H_3O^+]=(0.0729*g)/(36.46*g*mol^-1)xx1/(50xx10^-3L)#; this gives an answer in #mol*L^-1# as is required for a concentration.

And finally.....................

#pH=-log_10{(0.0729*g)/(36.46*g*mol^-1)xx1/(50xx10^-3L)}#

#=# #??#

What is #pOH# of this solution?