2. Write the balanced chemical equation and acid-equilibrium expression (Ka) for
ID: 560279 • Letter: 2
Question
2. Write the balanced chemical equation and acid-equilibrium expression (Ka) for the dissociatión of HC2H302 in water. Perform the necessary calculations to determine the expected pH of a 0.10 M solution of HC2H302. Show your work 3. Write the balanced chemical equation and base-equilibrium expression (Kb) for the dissociation of NaC2H302 in water. Perform the necessary calculations to determine the expected pH of a 0.10 M solution of NaC2H302. Show your work. 4. Experimentally, acetate buffer HC2H3O2-C2H302 can be prepared in different ways. Calculate the approximate pH of an acetate buffer solution prepared by mixing equal volumes of 0.10 M HCzHs02 and 0.10 M NaC:H,02 (Ka,HC2H302 = 1.8 x 105). Show your work. a. b. If an acetate buffer solution was going to be prepared by neutralizing HC2H302 with 0.10 M NaOH, what volume (in ml) of 0.10 M NaOH would need to be added to 10.0 mL of 0.10 M HC2H302 to prepare a solution with pH -5.5? Show your work.Explanation / Answer
2) Write down the dissociation of acetic acid, HC2H3O2 in water.
HC2H3O2 (aq) + H2O (l) ---------> H3O+ (aq) + C2H3O2- (aq)
The acid-dissociation constant is given as
Ka = [H3O+][C2H3O2-]/[HC2H3O2] (the concentrations of solid and liquid species aren’t included in the equilibrium constant expression)
= 1.8*10-5 (known from literature)
Let [H3O+] = x M; due to the 1:1 nature of dissociation, we must have [C2H3O2-] = x M and [HC2H3O2] = (0.10 – x) M. Put the values in the Ka expression.
Ka = (x).(x)/(0.10 – x)
===> 1.8*10-5 = x2/(0.10 – x)
Assume x is small; this is because Ka is small and hence, acetic acid mostly remains in the unionized form; therefore, we can write [HC2H3O2] = (0.10 – x) M 0.10 M.
1.8*10-5 = x2/(0.10)
====> x2 = 1.8*10-6
====> x = 0.0013416 0.00134
We have [H3O+] = 0.00134 M and pH = -log [H3O+] = -log (0.00134) = -(-2.8279) 2.83 (ans).
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