Calculate the pH at the following points on the titration curve where 10.00 mL of 0.250 M NH3 is being titrated with 0.100 M HCl. (0, 25, 50, 75, 100, 125% titrated)
Added by Craig O.
Step 1
00 mL of 0.250 M NH3 and 0.100 M HCl Since no HCl has been added yet, the solution is just NH3. NH3 is a weak base, so we need to calculate the pOH first. pOH = -log(Kw/Kb) = -log(1.0 x 10^-14 / 1.8 x 10^-5) = 4.74 pH = 14 - pOH = 14 - 4.74 = 9.26 Show more…
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