Calculate the pH of a solution that was created by combining 250.0 mL of 0.25 M HCl with 100.0 mL of 0.11 M HI. Assume the volumes are additive.
Both HCl and HI are strong acids and ionize completely; i.e. 100%
total volume = 250 + 100 = 350 mL = 0.350 L
mols HCl = M x L = 0.25 x 0.250 = ?
mols HI = 0.11 x 0.100 = ?
total mols = mols HCl + mols HI = ?
M of the solution = total mols/total volume in L = (H^+)
Then pH = -log (H^+)
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To calculate the pH of the solution, we need to determine the total concentration of H+ ions in the solution.
Step 1: Calculate the moles of H+ ions from HCl.
Moles of H+ ions from HCl = (volume of HCl solution in liters) x (molarity of HCl)
Given:
Volume of HCl solution = 250.0 mL = 0.250 L
Molarity of HCl = 0.25 M
Moles of H+ ions from HCl = 0.250 L x 0.25 M = 0.0625 moles
Step 2: Calculate the moles of H+ ions from HI.
Moles of H+ ions from HI = (volume of HI solution in liters) x (molarity of HI)
Given:
Volume of HI solution = 100.0 mL = 0.100 L
Molarity of HI = 0.11 M
Moles of H+ ions from HI = 0.100 L x 0.11 M = 0.011 moles
Step 3: Calculate the total moles of H+ ions in the solution.
Total moles of H+ ions = moles of H+ ions from HCl + moles of H+ ions from HI
= 0.0625 moles + 0.011 moles
= 0.0735 moles
Step 4: Calculate the final concentration of H+ ions in the solution.
Total volume of the solution = volume of HCl solution + volume of HI solution
= 0.250 L + 0.100 L
= 0.350 L
Final concentration of H+ ions = Total moles of H+ ions / Total volume of the solution
= 0.0735 moles / 0.350 L
≈ 0.210 M
Step 5: Calculate the pH of the solution.
pH = -log[H+]
pH = -log(0.210)
pH ≈ 0.68
Therefore, the pH of the solution is approximately 0.68.
To calculate the pH of a solution, we need to consider the properties of the acid and its concentration. In this case, we have two acids, HCl and HI, with different concentrations. We can find the resulting pH by following these steps:
1. Calculate the moles of HCl and HI using the given volumes and concentrations:
Moles of HCl = volume (in liters) × concentration of HCl
Moles of HI = volume (in liters) × concentration of HI
Volume HCl = 250.0 mL = 0.250 L
Concentration HCl = 0.25 M
Volume HI = 100.0 mL = 0.100 L
Concentration HI = 0.11 M
Moles of HCl = 0.250 L × 0.25 M = 0.0625 moles
Moles of HI = 0.100 L × 0.11 M = 0.011 moles
2. Determine the total moles of acid in the solution by summing the moles of HCl and HI:
Total moles of acid = moles of HCl + moles of HI
Total moles of acid = 0.0625 moles + 0.011 moles = 0.0735 moles
3. Calculate the molarity of the resulting solution by dividing the total moles of acid by the total volume in liters:
Total volume = volume of HCl + volume of HI
Total volume = 0.250 L + 0.100 L = 0.350 L
Molarity of the resulting solution = Total moles of acid / Total volume
Molarity of the resulting solution = 0.0735 moles / 0.350 L ≈ 0.210 M
4. Use the molarity of the resulting solution to calculate the pOH using the formula: pOH = -log10(Molarity of the resulting solution)
pOH = -log10(0.210) ≈ 0.6776
5. Calculate the pH by subtracting the pOH from 14 (pOH + pH = 14):
pH = 14 - pOH
pH = 14 - 0.6776 ≈ 13.3224
Therefore, the pH of the solution created by combining 250.0 mL of 0.25 M HCl with 100.0 mL of 0.11 M HI is approximately 13.3224.