00:01
To calculate the concentrations of all species in solution of a .490 molar h2s .3 solution, we'll first write the two equilibrium that can occur for ka1 and ka2.
00:15
We'll then make the assumption that ka1, being so much larger, is the only significant contributor to the hydronium ion concentration.
00:24
In addition, with it being the only significant contributor and the stoichiometry one -to -one, we will assume that these two concentrations, h .s .o .3 minus, and the hydronium concentration are equal to each other.
00:39
Then we can begin solving for the hydronium concentration just using the k -a -1 expression.
00:45
The k -a -1 expression can be rewritten as 1 .6 times 10 to negative 2 is equal to the hydrogen concentration squared divided by the initial concentration of h2s -o3 minus the extent to which it goes to the right so minus the hydrogen concentration then using our calculator or excel or mathematica or long hand in the quadratic formula whatever your instructor has advised you to do you'll solve for the hydrogenum concentration the hydrogen concentration based on our assumptions will also be equal to the h.
01:22
S .o .3 minus concentration, and from this equation, we get 0 .0809 molar.
01:30
The h2s .03 minus concentration will be the amount we start with minus the hydrogenium concentration, or 0 .409...