A 172.8mL sample of 0.360M acetic acid (CH3CO2H) was allowed to react with 1.835

Question

A 172.8mL sample of 0.360M acetic acid (CH3CO2H) was allowed to react with 1.835L of gaseous ammonia at 25 degrees C and a pressure of 630.0mm Hg. Assuming no change in the volume of the solution, calculate the pH. Values of equilibrium constants are: Acetic Acid: CH3CO2H: Ka=1.8 * 10^-5 Ammonia: NH3: Kb=1.8*10^-5 please help!

Explanation / Answer

CH3COOH ---> CH3COO- + H+ 0.36-x x x Ka = 1.8 x10^ -5 = x^2/(0.36-x) x = [H+] = 0.002578 moles of H+ = 0.002578 x172.8/1000 =0.00044548 NH3 + H2O---> OH- + NH4+ PV = nRT , (630/760) x1.835 = n x0.0821 x298 n = 0.06217 mole s , [NH3] = 0.06217/1.835 = 0.03388 Kb = 1.8 x10^ -5 = x^2/(0.03388-x) x = 0.000772= [OH-] moles of OH- = 0.000772 x1.835 = 0.001416 net OH- = 0.001416-0.00044548 = 0.00097 vol = (0.1728+1.835) = 2.0078 [OH-] = 0.00097/2.0078 =0.000483 pOH = 3.316 pH = 14-3.316 =10.684

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