DrBob has given you the equation to use:
k2 = [H^+][SO4^-]/[HSO4^-]
and the values to substitute
[H^+] = 0.01 + x
[SO4^-2] = x
[HSO4^-] = 0.01 - x
k2 = 1.2 x 10^-2
you need to substitute these and solve for x.
Remember that you are after 0.01+x [H^+]
to calculate the pH.
is the answer ph of 2.398
3 answers
I don't obtain that for an answer.
Nor do I. The first ionization of concentration .01 will give a pH of 2, so adding more H from the second ionization has to make the pH more acid (ie less than 2).
In one of your earlier posts, you asked if this was the correct quadratic equation and it was.
X^2 + 0.022X - 1.2E-4 = 0
Using the quadratic formula, which is
X=(-b +/- sqrt[b^2-4ac])/2a
solve for X. What is that value?
X^2 + 0.022X - 1.2E-4 = 0
Using the quadratic formula, which is
X=(-b +/- sqrt[b^2-4ac])/2a
solve for X. What is that value?
2 answers