Asked by lyne
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
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
Answers
Answered by
DBob222
I don't obtain that for an answer.
Answered by
bobpursley
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).
Answered by
DBob222
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?
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