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### Chapter 12 problem 79

Posted: Fri Dec 02, 2016 8:57 pm
12.79
Calculate the pH of 0.15 m H 2 SO 4 (aq) at 25 C.

### Re: Chapter 12 problem 79

Posted: Sat Dec 03, 2016 2:28 pm
Sulfuric acid is polyprotic so there are two deprotonizations. The first time, H2SO4 is completely ionized into HSO4- and H3O+. Since the ionization is complete, the concentration of HSO4- is 0.15M and the concentration of H3O+ is 0.15M (same concentration as H2SO4). The next ionization is incomplete, so an ICE table is needed. The equation for the second ionization is HSO4- + H2O (double arrow) SO42- + H3O+. The initial concentrations are taken from the 1st ionization, so [HSO4-]= 0.15M, [SO42-]=0, and [H3O+]=.15M. Set up the Ka as usual so it's (0.15+x)(x)/(0.15-x)= 1.2x10^-2. Use the second Ka value for H2SO4 which can be found on pg. 496 in the textbook. Ka is too large for x to be ignored, therefore use the quadratic formula to solve for x. X=0.0104M. 0.15+x is the equilibrium concentration of hydronium ions so the 0.15+0.0104=0.16M [H3O+]. Taking the negative log of that gives pH=0.80

### Re: Chapter 12 problem 79

Posted: Sun Dec 04, 2016 11:11 am
Thank you! Your explanation is very helpful! I was also confused on how to do this problem!