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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.
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Answered by
DrBob222
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^+)
Post your work if you get stuck.
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^+)
Post your work if you get stuck.
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