HC12H17ON4SCl2 is a weak acid with Ka=3.4 x 10-7. Suppose a 0.250F of thiamine h

Question

HC12H17ON4SCl2 is a weak acid with Ka=3.4 x 10-7. Suppose a 0.250F of thiamine hydrochloride is dissolved in water. Use a systematic treatment of equilibrium to determine the pH and fraction of dissociation. Ignoring activity.

Suppose a 0.250M of thiamine chloride is dissolved in water. Use a systematic treatment of equilibrium to determine the conjugate base concentration at pH=5. Ignoring activity.

How to solve systematic treatment of equilibrium questions? It is hard to figure out how to set charge balance and mass balance and combine equations to solve problem..

Thank you!!

Explanation / Answer

Represent HC12H17ON4SCl2 as HA. HA undergoes dissociation to give

HA--à H+ + A-

[A-] is conjugate base

Ka =[H+] [A-]/[HA]

Initial concentrations : [HA]= 0.25M , [H+] =[A-] =0

Let x = equilibrium concentration of [H+] =[A-]

At equilibrium [HA]= 0.25-x

Ka= 3.4*10-7= x2/(0.25-x)

Based on the magnitude of Ka 0.25-x can be approximated to 0.25

Hence x2/0/25= 3.4*10-7, x2=3.4*0.25*10-7=8.5*10-8,   x=0.00029 M =[H+] =0.00029

pH= -log(0.00029)=3.535

degree of dissociation can be calculated based on amount dissociated which is 0.00029

and amount present = 100*0.00029/0.25=0.116%

at PH=5 , -log[H+] =5 [H+] =10-5 =[A-]

conjugate base concentration =10-5M

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