If 0.03 moles of sulfuric acid is mixed
with water to make 266 mililiter of solution, what is the molarity of H+?
Answer in units of M.
How advanced is this course? This isn't the simple problem it appears.
M H2SO4 = moles/L = 0.03/0.266 = ??
H2SO4 ==> H^+ + HSO4^-
0.113M...0.113...0.113
HSO4^- ==> H^+ + SO4^2-
0.113-x....x......x
k2 for H2SO4 = about 0.012 = (H^+)(SO4^2-)/(HSO4^-)
Substitute into k2 expression and solve for x. I get 0.01 for x which means H^+ = 0.123M (not twice the molarity as one might think).
You should confirm k2 and all of the work. The k2 expression evolves into a quadratic OR you may solve it by iteration.
H2SO4 ==> H^+ + HSO4^-
24345
To find the molarity of H+, we need to calculate the concentration of H+ ions in the solution.
First, we need to determine the number of moles of H+ ions present in the solution. Sulfuric acid (H2SO4) is a strong acid, which means it completely dissociates in water to produce two H+ ions for every molecule of H2SO4.
Since we have 0.03 moles of sulfuric acid, we would have twice as many moles of H+ ions. Therefore, the number of moles of H+ ions is 2 * 0.03 mol = 0.06 mol.
Next, we need to calculate the volume of the solution in liters. We know that 1 mL is equal to 0.001 L. So, 266 mL is equal to 266 * 0.001 L = 0.266 L.
Finally, we can calculate the molarity (M) of H+ by dividing the number of moles of H+ ions by the volume of the solution in liters:
Molarity (M) = moles of solute / volume of solution in liters
Molarity (H+) = 0.06 mol / 0.266 L
Calculating this gives us the molarity of H+ as approximately 0.226 M (rounded to three decimal places).