In the titration of acetic acid (HAcO, K_a = 1.8 times 10^-5) with NaOH the reac
ID: 955073 • Letter: I
Question
In the titration of acetic acid (HAcO, K_a = 1.8 times 10^-5) with NaOH the reaction is HAcO(aq) + OH^-(aq) leftarrowrightarrowAcO^-(aq) + H_2O(liq). What is the value of the equilibrium constant of the reaction? 1.8 times l0^-5 5.6 times l0^4 5.6 times l0^-10 1.0 times l0^-14 1.8 times l0^9 The ionization constant of a certain acid base indicator HIn is K_a(HIn) = 2.3 x 10^-5. The color of the deprotonated form of the indicator in solution is blue, while the color of the deprotonated form of the indicator in solution is colorless. Which one of the following statements is (are) incorrect? A solution with the indicator changes color in the pH range 3.6 - 5.6. The indicator is unsuitable for the titration of acetic acid, HAcO, with NaOH. The sample solution is colorless when its pH is 2.0. The sample solution is light blue when its pH is 8.6. In a titration the indicator is dissolved in the titrant solution. (1) and (3) (2), (3), and (5) (4) and (5) (5) (3), (4), and (5)Explanation / Answer
HAcO + H2O (l) -------> H+ + AcO-
Ka = [H+] [AcO- ] / [ HAcO] = 1.8 * 10-5
AS the reaction is given by
HAcO + OH- -------> H2O(l) + AcO-
K = [AcO- ] / [ HAcO] [OH- ] .....................equation 2
As [H+ ] [OH-] = 10-14 or [OH-] = [H+ ] / 10-14
and putting in equation 2
K = [AcO- ] 10-14 / [ HAcO] [H+] ........................ equation 3
putting value of (Ka= [AcO- ] / [ HAcO] [H+] ) in equation 3
K = Ka * 10-14 = 1.8 * 10-5 * 10-14
K = 1.8 * 10-9
the correct anser is D)
2) the incorrect state ment is
3rd the sample solution is colourless when the pH is 2.0
As the Ka(HIn) = 2.3 * 10-5
as (HIn) ---> H+ + In-
H+ = 2.3 * 10-5
and pH= 4.63
as the question said that the colour of the protonated form of indicator in solution is blue this means that the solution has more concentartion of H+ and the solution is acidic in nature
but the 3rd option said that at ph = 2 the solution the colour is colourless
this is a false statement .
4) th is incorrect as the pH= 8.6 this means the solution has less concentration of H+ and as the question says that the deprotonated solution must be colourless then how this solution get coloured
4 th is also incorrect
5) th is also incorrect because the indicator is dissolved in the analyte. not in the titrant .
E) is the correct option .