Question

1) Write the chemical reaction for hydrogen thiocyanate in water, whose equilibrium constant is ?a. Include the physical states for each species.

?a reaction:

Write the chemical reaction for the thiocyanate ion in water, whose equilibrium constant is ?b. Include the physical states for each species.

?b reaction:

2)

Enough of a monoprotic acid is dissolved in water to produce a 1.24 M solution. The pH of the resulting solution is 2.84 Calculate the K_a for the acid.

?a=

3)

Ascorbic acid (H2C6H6O2) is a diprotic acid with ?a1=8.00×10−5 and ?a2=1.60×10−12. Determine the pH of a 0.303 M ascorbic acid (H2C6H6O2) solution.

pH=

Answer 1

1) HSCN (aq) + H2O (l) --------> SCN- (aq) + H3O+ (aq)

Ka = [SCN-] [H3O+] / [HSCN]

SCN- (aq) + H2O (l) ----------> HSCN (aq) + OH- (aq)

Kb = [HSCN] [OH-] / [SCN-]

2) HA (aq) + H2O (l) ------------> A- (aq) + H3O+ (aq)

1.24 M 0 0 (at t =0)

1.24 - x x x (at t = equilibrium)

But x = [H3O+] =10^-pH = 10^(-2.84) = 0.00145M

Since x is very less so, we can consider 1.24-x = 1.24

So, Ka = [H3O+] [A-] / [HA] = (0.00145)^2/1.24 = 1.7*10^-6

Ka = 1.7*10^-6 ......... Answer

3) H2C6H6O2 (aq) -------------> 2H+ (aq) + C6H6O2--(aq)

0.303 M 0 0 (at t=0)

0.303-x 2x x (at t =equilibrium)

Final Ka = Ka1 * Ka2 = 8*10^-5 * 1.6*10^-12 = 12.8*10^-17

Ka = (2x)^2 * x / 0.303 ( since x is very small )

1.28*10^-16 = 4x^3/ 0.303

x^3 = 1.28*10^-16*0.303 / 4 = 10^-17

x = 2.2*10^-6

[H+] = 2x = 4.4*10^-6

pH = -log [H+] = - log (4.4*10^-6) = 5.357 ...... Answer