help with homework 12.57

[905084274]

can someone explain step by step how you solve 12.57 6th edition. I think the solutions manual is incorrect or is just confusing...

[Madeline Ho 1C]

For part a, you're given that the pH of 0.10 M HCLO2 is 1.2. pH is -log[H3O+], so to solve for [H3O+] you have to do 10^-pH. In this case, it would be 10^-1.2=0.06 M H3O+. Then, you solve for the Ka (acidity constant), which is (0.06^2)/(0.1-0.06). The pKa is the -log of your Ka value. Part b is the exact same process except [OH-] because C3H7NH2 is a base. Hope that helps!

[Mukil_Pari_2I]

For part a, you're given that the pH of 0.10 M HCLO2 is 1.2. pH is -log[H3O+], so to solve for [H3O+] you have to do 10^-pH. In this case, it would be 10^-1.2=0.06 M H3O+. Then, you solve for the Ka (acidity constant), which is (0.06^2)/(0.1-0.06). The pKa is the -log of your Ka value. Part b is the exact same process except [OH-] because C3H7NH2 is a base. Hope that helps!

Do we also only use 1 sf throughout the calculation bc the pH has one sf or do we do take into account the sfs at the end?