What is the hydrogen ion concentration, in mol dm−3, when 2.00 g of sulfuric acid is dissolved in water to give 0.400 dm3 of solution?
How advanced is this class? I suppose the beginning class you would expect
mols H2SO4 = grams/molar mas = about 0.0510. Two mols H^+/mol H2SO4 = (H^+) = 2*0.0510. However, in reality, the second H in H2SO4 is not a strong acid as is the first one so it contributes only a minimal amount of H^+ and the (H^+) actually is 0.0510 from the first H + 0.00855 from the second. Add them for the total (but watch the number of significant figures). I can show you how to do the more advanced calculation if you are to go that route.
Thanks- I think I've cracked it x
To find the hydrogen ion concentration in the given solution, we need to calculate the number of moles of hydrogen ions (H+) present and then divide it by the volume of the solution.
To start, we need to determine the number of moles of sulfuric acid (H2SO4) in the solution. We can use the formula:
moles = mass / molar mass
The molar mass of sulfuric acid (H2SO4) is:
(2 * atomic mass of hydrogen) + atomic mass of sulfur + (4 * atomic mass of oxygen)
Atomic masses:
Hydrogen (H): 1.01 g/mol
Sulfur (S): 32.07 g/mol
Oxygen (O): 16.00g/mol
Calculating the molar mass:
(2 * 1.01) + 32.07 + (4 * 16.00) = 98.09 g/mol
Now, we can calculate the number of moles of sulfuric acid:
moles = 2.00 g / 98.09 g/mol ≈ 0.0204 mol
Next, we divide the moles of sulfuric acid by the volume of the solution to find the hydrogen ion concentration:
Hydrogen ion concentration (mol dm^-3) = moles / volume
Hydrogen ion concentration = 0.0204 mol / 0.400 dm^3 ≈ 0.051 mol dm^-3
Therefore, the hydrogen ion concentration is approximately 0.051 mol dm^-3.