Question

What will be the equilibrium concentration (in molarity) of the sulfate ion if the prepared concentration of sulfuric acid was 0.015 M instead of 0.031 M? (Hint: Use the quadratic formula. Also, use the same Ka1 and Ka2 values from the previous problem.)

a) 0.031 M

b) 0.015 M

c) 0.045 M

d) 0.062 M

Answer 1

The equilibrium concentration of the sulfate ion (SO4²-) will be 0.015 M.

Step-by-step explanation:

To calculate the equilibrium concentration of the sulfate ion (SO4²-) when the concentration of sulfuric acid (H2SO4) is 0.015 M instead of 0.031 M, we can use the quadratic formula. Let's assume the equilibrium concentration of SO4²- is x. The initial concentration of H2SO4 is 0.015 M, so the concentration of H3O+ (which is equivalent to the concentration of H2SO4) is also 0.015 M.

Using the given Ka values, we can write the equilibrium expression for the dissociation of H2SO4 as: Ka1 = ([H3O+][HSO4-])/[H2SO4] = (0.015 * x)/(0.015) = x.

Simplifying the equation, we get: x² = 0.015x. Rearranging, we have: x² - 0.015x = 0. Solving this quadratic equation using the quadratic formula, we find that x = 0.015 M or x = 0.

Since we're looking for the equilibrium concentration of SO4²-, we discard the value of x = 0. Therefore, the equilibrium concentration of the sulfate ion is 0.015 M.