Answers
![[salt] pH = pka + log [ Acid the get of Now, Here given, TAcid) = CH2, Cro H = 0-4 M [salt] = [CH3COONa] = 0.2 m for acetic a](http://img.homeworklib.com/questions/eacafdf0-b800-11eb-b073-b9f04765ce92.png?x-oss-process=image/resize,w_560)
![for these type of Solution - [salt] рон ерко тоГоле) Kb = ionisation constant of weak base Again, pH = 14 -POH Now given - [B](http://img.homeworklib.com/questions/ec1a6680-b800-11eb-8da3-c9dcbbc43c6d.png?x-oss-process=image/resize,w_560)
Please,,,,,for the other answers post them as another questions.....
& Finst questions : Determination ut ut i - Here the given acid is acetic acid (CHg. CooH). It is a weak acid. For any weak acid the concentration of Hgot ion is - [H30 t] - [ ka x = (exka Again by definition - pH = -log [ttso t] = - 12 log Ka - loge a pH = 2 pka - 22 loge [ipka = -log ka] ka=ionisation constant of acid c = concentration of weak acid.
Now given that I the concentration c = 0.4 M And. pka of CH2 LOUH = 4.76 (data collected). Now from equation ③ we have - pH = 1 (4.76) – 12 log 0.4 = 2.38 – 0.2 = 2.58 Now, the pH of 0.4M CH3COH solution = 2.58 (Aus) - Acid => CH3 COOH (weak). Salt = CHg cogna So, it is a solution of weak acid and its salt, so it acts as a buffers solution Now, for buffer solution the expression of pt can be given as
[salt] pH = pka + log [ Acid the get of Now, Here given, TAcid) = CH2, Cro H = 0-4 M [salt] = [CH3COONa] = 0.2 m for acetic acid, - O pka = 4.76. Now, froom eqr 2 we can pH of the solution as - pH = 4.76 + log Ich y [CHg cook = 4.76 + logo = 4.76 + (-0.3) = 4.46 & pH of the given solution = 4.46 fans) [CH₂ Loona 0.2
(Reference: Change 16-5) A buffen solution is the mixture of solutions of weak acid and its salt.
Now HU and Hq so both are strong acid. So both HU and Hq So do not form buffes solution, so options @ and are ruled out. Again, Nitrous acid HNO 2 Weak acid Potassium nitrite KNO2 Salt Mixture of HNO and KNO2 can act as a buffers So the connect option is - (c) KNO2/HNO2 (Reference change 16.9) A mixture of weak base and its salt can also act as a buffer I Solution.
for these type of Solution - [salt] рон ерко тоГоле) Kb = ionisation constant of weak base Again, pH = 14 -POH Now given - [Base] = [Nitg] = 0.15 m Too NHg act as a weak base] [salt] - [Nitqul] = 0.35 m a. From equation ④ - pOH = pkb + log - [Nitqu Log In Ho] lopky of NH3 -4.757 a pot = 4.75 + log 0.35 o 0.15 = 4.75 +0.37 = 5.12 from equation 2 pH = 14-5.12 = 8.88 Ans
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