What is the ph of a 1.0 x 10^-5 m solution of sulfuric acid (h2so4)?

   The PH  of  1.0 x10^-5  solution  of sulfuric acid(H2SO4)  is   4.7

calculation

PH =-log (H^+)
to calculate (H^+)  concentration
write the equation for dissociation of  H2so4

H2SO4⇔  2H^+  + SO4^2-

  if the concentration  of H2SO4 = 1.0 x10^-5  and the  mole  ratio  between H2SO4   to H^+ is   1:2  therefore    the  concentration of H^+ 
= 2 x(1.0 x10^-5)= 2 x10^-5 M

therefore Ph   =- log  (2x10^-5)  =  4.7

It shows that the experiment is designed properly.