A 1.0 M H¿S solution has a pH = 3.75 at equilibrium. What is the value of Ka?

K_a = ( [H⁺] [A¯] ) / [HA] ( our dissociation equation set up )

So, H^+ = 10^-pH

10^ - 3.75

= 1.778 X 10^-4

------A^- is one to one with H^+ and we are given HA.

Ka = (1.778 X 10^-4)(1.778 X 10^-4)/1.0 M

= 3.2 X 10^-8

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