My gut feeling is that the author intended this to be a hydrolysis problem but I don't think it's that simple. As a simple hydrolysis problem th equation is
............HCO3^- + HOH ==> H2CO3 + OH^-
I.............0.240.........................0.................0
C...............-x............................x...................x
E...........0.240-x........................x...................x
Kb for HCO3^- = (K2/Ka for H2CO3) = (x)(x)(-/240-x)
Solve for x = (OH^-)
However, this is a more complicated problem than that. Not only does the HCO3^- hydrolyze as above BUT that is complicated by another reaction that is HCO3^- + H2O ==> H3O^- + CO3^- for which you can write a second hydrolysis equation. The net effect is that HCO3^- sits in the middle of the H2CO3/HCO3^-/CO3^2- and can go both ways. In the case of HCO3^, the pH is determined by pH = (pK1 + pK2)/2 or (H^+) = sqrt(k1*k2) for intermediate strengths of HCO3^-.
Determine the [OH−] of a solution that is 0.240 M in HCO3− (Ka1=4.3×10−7).
Determine the pH of this solution.
4 answers
how do you solve for x
do we use quadratic formula?
I don't know for sure but I suspect you will need to use the quadratic formula. The answer using the quadratic formula will be your best bet.
1 answer