The p H at the equivalent point for the titration of 0.10 M KH 2 BO 3 with 0.1 M Hl is ( mathrmK mathrma. of . mathrmH3 mathrmBO3=12.8 × 10-10) Report your answer by rounding it up to nearest whole number.

Question

The p H at the equivalent point for the titration of 0.10 M KH 2 BO 3 with 0.1 M Hl is ( mathrmK mathrma. of . mathrmH3 mathrmBO3=12.8 × 10-10) Report your answer by rounding it up to nearest whole number. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

First there will be an acid-base reaction between KH2BO3 and HCl . Solution

conc. will become half after reaction of an equal volume of K H2B O3 and HCl So (0.1/5)=0.05 Now, Solution

[H+]=√Ka C=√12.8 × 10- 10 × 0.05=8× 10- 6 mathrmpH=- log (8 × 10-6) ≈ 5.0

Q. The pH at the equivalent point for the titration of 0.10MKH2​BO3​ with 0.1MHl is (Ka​ of H3​BO3​=12.8×10−10) Report your answer by rounding it up to nearest whole number.

First there will be an acid-base reaction between KH2​BO3​ and HCl . conc. will become half after reaction of an equal volume of KH2​BO3​ and HCl So 50.1​=0.05 Now, [H+]=Ka​C​=12.8×10−10×0.05​=8×10−6 pH=−log(8×10−6)≈5.0

Q. The $p H$ at the equivalent point for the titration of $0.10 M K H_{2} B O_{3}$ with $0.1 M H l$ is $(K_{a}$ of $H_{3} BO_{3} = 12.8 \times 1 0^{- 10})$ Report your answer by rounding it up to nearest whole number.