I must admit this is a new one on me; however, I assume we have HClO at the equivalence point.
moles NaClO = 1.86/about 74 = about 0.025
moles HClO = 0.025 at equivalence point.
How much 1M HCl is required to titrate the sample? That will be M = mmols/mL or
mL = mmols/M = 25 mmoles/1M = 25 mL. The initial volume was 250 mL so the final volume is 250+ 25 = 275 mL.
(HClO) = mmoles/total volume = 25/275 = 0.0909M
..........HClO ==> H^+ + ClO^-
initial...0.0909....0......0
change......-x.......x......x
equil.....0.0909-x...x.......x
Ka = (H^+)(ClO^-)/(HCl0)
Substitute from the ICE chart above into the Ka expression and solve for (H^+), then convert to pH by pH = -log(H^+).
Post your work if you need additional assistance.
Sodium hypochlorite, NaClO, is added as a disinfectant to the water supply. You
are analyzing a sample from the supplier to verify its purity and you prepare to titrate 250.0 mL of a solution containing 1.86 g of NaClO with a 1.00 M HCl solution.
What is the pH of the solution at the equivalence point?
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