After 10 ml of added base Moles base added = 0.138*0.005 = 0.00069 This equals moles of acid reacted Moles acid before reaction = 0.025*0.11 = 0.00275 Moles acid remaining = 0.00275 - 0.00069 = 0.00206Volume = 25 + 10 = 0.035L Concentration acid remaining = 0.0021/0.035 = 0.06M If x moles dissociate: 0.06 - x = x + x [H+][OAc-]/[HOAc] = x*x/(0.06-x) = 1.76*10^-5 Assume x <<< 0.06, then 0.06-x ~ 0.06 x^2 = 0.06*1.74*10^-5 x = [H+] = 1.02*10^-3pH = 3.0....Do the same steps for 20mL and 30mL
it says the answer is wrong :(
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