IV. Acid-Base Titration (15 points). A 50 mL of 0.200 M HNO2, (Ka= 4.0 x 10)...

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Answer 1

Answer #1

Answer:-

Firstly ionization equation is written then using the expression of Ka and its value, pH is calculated at different points during titration.

The answer is given in the image,

Answer:- CHNO27= 0.200 kq -4.0x104 CHQOH] =0-200M 29) before tit ration HNO2 +420 H3O+ + NO₃- 0.200-X Ka = [4307][NO] [HNOz]

so (Ab04) Kq = 4.0X10-4 PH=- log (4.0X10-4) - PH=3.40 C4 Saml NaOH added, It is escuedence point So [NO2] = 0.200 = 0.100M NO

Answer 2

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