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
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?
To solve for x and find the pH, we can substitute the given values into the equation:
k2 = [H^+][SO4^-]/[HSO4^-]
Substitute the known values:
1.2 x 10^-2 = (0.01 + x)(x)/(0.01 - x)
Now, let's simplify the equation:
(0.01 + x)(x) = (1.2 x 10^-2)(0.01 - x)
Expand the left side:
0.01x + x^2 = 1.2 x 10^-4 - 1.2 x 10^-2x
Combine like terms:
x^2 + 0.01x - 1.2 x 10^-4 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
Applying the quadratic formula to our equation:
x = (-0.01 ± √((0.01)^2 - 4(1)(-1.2 x 10^-4)))/(2(1))
Simplifying inside the square root and solving, we get two solutions for x:
x ≈ -0.01199 or x ≈ 0.00099
Since a negative value for [H^+] doesn't make sense in this context, we discard the negative value. Therefore, x ≈ 0.00099.
Now that we have the value of x, we can find the pH by adding it to 0.01:
pH = 0.01 + 0.00099 ≈ 0.01099
So, the pH is approximately 0.01099, not 2.398.