The acid dissociation constant of #"H"_2"S"# and #"HS"^-# are #10^-7# and #10^-13# respectively. The pH of 0.1 M aqueous solution of #"H"_2"S"# will be?

To find the #pH# we must find how far it has dissociated:
Let's set up some equation using the #K_a# values:

#K_a(1)=([H_3O^+] times [HS^-])/([H_2S])#

#K_a(2)=([H_3O^+] times [S^(2-)])/([HS^(-)])#

This acid will dissociate in two steps. We are given the concentration of #H_2S# so lets start from the top and work our way down.

#10^-7=([H_3O^+] times [HS^-])/([0.1])#

#10^-8=([H_3O^+] times [HS^-])#
Then we can assume that both of these species are in a 1:1 ratio in the dissociation, allowing us to take the square root to find the concentration of both species:

#sqrt(10^-8)=10^-4=([H_3O^+] = [HS^-])#

Now in the second dissociation, #[HS^-]# will act as the acid. That means we plug in the concentration found in the first calculation in the denominator of the second dissociation:

#10^-13=([H_3O^+] times [S^(2-)])/([10^-4])#

Same principle to find the concentration of #[H_3O^+]#:

#10^-17=([H_3O^+] times [S^(2-)])#

Hence:
#sqrt(10^-17)=3.16 times 10^-9=[H_3O^+] = [S^(2-)]#

So the combined concentration of #H_3O^+# will be:
#10^-4 + (3.16 times 10^-9) approx 10^-4#

#pH=-log[H_3O^+]#
#pH=-log[10^-4]#
#pH=4#

So the second dissocation was so small it did not really impact the pH. I guess if this was a multiple choice exam then you only needed to look at the first dissociation and find the square root of #10^-8# to find the #H_3O^+# concentration, and hence the #pH# using the log law:

#log_10(10^x)=x#

But it is always good to be thorough :)