The curve for the titration of 50.0 mL of 0.0200 M HClO(aq) with 0.100 M NaOH(aq

Question

The curve for the titration of 50.0 mL of 0.0200 M HClO(aq) with 0.100 M NaOH(aq) is given below. Estimate the pK_a of HCIO The pK_b for CH_3NH_2 is 3.44. This expression refers to which of the following reactions? CH_3NH_3^+(aq) + H_20(l) doubleheadarrow CH_3NH_2(aq) + H_3O^+(aq) CH_3NH_3^+(aq) + OH^-(aq) doubleheadarrow CH_3NH_2(aq) + H_2O(aq) CH_3NH_2(aq) + H_3O^+(aq) doubleheadarrow CH_3NH_3^+(aq) + H_2O(l) H_3O^+(aq)+- OH^-(aq) doubleheadarrow 2H_2O(I) CH_3NH_2(aq) + H_2O(l) doubleheadarrow CH_3NH_3^+(aq) + OH^-(aq) The pK_a for HF is 3.45. This expression refers to which of the following reactions? HF(aq) + H_2O(I) doubleheadarrow F^-(aq) + H_3O^+(aq) HF(aq) + OH^-(aq) doubleheadarrow F^-(aq) + H_2O(l) H_3O^+(aq) + OH^-(aq) doubleheadarrow 2H_2O(l) F^-(aq) + H_2O(l) doubleheadarrow HF(aq) + OH^-(aq) F^-(aq) + H_3O^+(aq) doubleheadarrow HF(aq) + H_2O(l) If the value of K_a for HF is 3.5 Times 10^-4, calculate the equilibrium constant for F^-(aq) + H_2O(l) doubleheadarrow HF(aq) + OH^-(aq) 2.9 Times 10^3 -3.5 Times 10^-4 3.5 Times 10^-18 2.9 Times 10^11 3.5 Times 10-4

Explanation / Answer

26.

pKa must be the pH of the half equivalence point

then

identify V of equivalence (when pH changes drastically)

V = 10 ml approx

then half way is 10/2 = 5 ml

the pH at 5 ml i sbetween 7 and 7.5

it is most near to 7.5 so choose 7.5

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