A) this is about the equivalence point for the titration of31.0 mL of 0.263 M pi

Question

A) this is about the equivalence point for the titration of31.0 mL of 0.263 M piperidine(pKb = 2.89) with 0.153 M HI. Calculate the pH of the solutionat the equivalence point.

B) this is about theequivalence point for the titration of 32.1 mL of0.340 M hypobromous acid (pKa =8.64) with 0.193 M KOH. Calculate the pH of the solutionat the equivalence point.
Calculate the pH of the solutionat the equivalence point.

B) this is about theequivalence point for the titration of 32.1 mL of0.340 M hypobromous acid (pKa =8.64) with 0.193 M KOH. Calculate the pH of the solutionat the equivalence point.
Calculate the pH of the solutionat the equivalence point.

B) this is about theequivalence point for the titration of 32.1 mL of0.340 M hypobromous acid (pKa =8.64) with 0.193 M KOH. Calculate the pH of the solutionat the equivalence point.

Explanation / Answer

at the equivalence point all of piperidine base reacts. moles of base = moles of acid added (B/c 1:1stoichiometry) B + HI BH+ + I-, where B is base, BH+ is protonated base(look ammonium cation). 31.0 mL *1L/1000mL * (0.263 mol/L) = 0.153 mol/L * V = 0.008153moles V = 53.3 mL of acid solution added. Initially there is 0.008153 moles/(53.3+31.0) mL *(1000mL/1L) = 0.0967 M of BH+ formed BH+ + H2O B + H3O+ This is described by pKa = 14 - pKb = 14 - 2.89 = 11.11; Ka =10^-11.11 Ka = [B][H3O+]/[BH+] [BH+] = initial - change = 0.0967 - x [H3O+] = x [BH+] = x Ka = 10^-11.11 = x^2/(0.0967 - x) Assume 0.0967 >> x 10^-11 = x^2/0.0967 x = 9.83 e-7 pH = -log[H3O+] = -log(x) = 6.01

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