A 150ml solution of 0.340M HCl is mixed with a solution of 340ml of 0.175M HNO3. The solution is then diluted to a final volume of 1.00L. How many moles of H+ are present in the final solution?
mols HCl = M x L = ?
mols HNO3 = M x L = ?
Total mols H^+ (from the two) = just add them. The part about the dilution has no effect. It affects the CONCENTRATION OF H^+ but not the number that are there.
To determine the moles of H+ present in the final solution, we need to calculate the moles of H+ from each component and add them together.
First, we calculate the moles of H+ from the HCl solution:
Moles of HCl = concentration (M) × volume (L)
Moles of HCl = 0.340 M × 0.150 L = 0.051 moles of H+
Next, we calculate the moles of H+ from the HNO3 solution:
Moles of HNO3 = concentration (M) × volume (L)
Moles of HNO3 = 0.175 M × 0.340 L = 0.0595 moles of H+
Now, we add the moles of H+ from each component to find the total moles of H+:
Total moles of H+ = moles of H+ from HCl + moles of H+ from HNO3
Total moles of H+ = 0.051 moles + 0.0595 moles = 0.1105 moles of H+
Therefore, there are 0.1105 moles of H+ present in the final solution.