A weak acid, Ha, is 0.1% ionized in a 0.2M solution. Answer the following questi
ID: 24182 • Letter: A
Question
A weak acid, Ha, is 0.1% ionized in a 0.2M solution. Answer the following questions:a) What is the equilibrium constant (Ka) for the dissociation of this acid?
b) What is the pH of this solution?
c) How much 'weaker' in active acidity of the HA solution compared to 0.2M HCl solution?
d) How many mls of 0.1 M NaOH would be needed to neutralize completely 250 ml of the 0.2M HA solution?
Explanation / Answer
a) HA --> H(+) + A(-) 0.1% ionized means 0.001 of initial [HA] is H+ and A- Ka = [H+][A-]/[HA], where [HA] is current concentration Ka = (0.001*0.2)(0.001*0.2)/(0.998*0.2) = 2.004 x 10^-7 b) pH = -log([H+]) = -log(0.001*0.2) = 3.70 c) % ionization of HCl = 100% Therefore 0.1%/100% = 0.001 ==> HA solution is 1000 times WEAKER than HCl d) 0.2M*0.25 = 0.05 mol HA --> 0.00005 mol H+ (.1% ionization) 0.00005 mol NaOH is needed. So: (0.1 M)*V = 0.00005 mol ==> V = 0.0005 L or 0.5 mL