A Buffer Solution Consists Of NH3 And NH4Cl Total Concentration ...

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AnswerQuestion Answers for Class 12Class 12 BiologyClass 12 ChemistryClass 12 EnglishClass 12 MathsClass 12 PhysicsClass 12 Social ScienceClass 12 Business StudiesClass 12 EconomicsQuestion Answers for Class 11Class 11 EconomicsClass 11 Computer ScienceClass 11 BiologyClass 11 ChemistryClass 11 EnglishClass 11 MathsClass 11 PhysicsClass 11 Social ScienceClass 11 AccountancyClass 11 Business StudiesQuestion Answers for Class 10Class 10 ScienceClass 10 EnglishClass 10 MathsClass 10 Social ScienceClass 10 General KnowledgeQuestion Answers for Class 9Class 9 General KnowledgeClass 9 ScienceClass 9 EnglishClass 9 MathsClass 9 Social ScienceQuestion Answers for Class 8Class 8 ScienceClass 8 EnglishClass 8 MathsClass 8 Social ScienceQuestion Answers for Class 7Class 7 ScienceClass 7 EnglishClass 7 MathsClass 7 Social ScienceQuestion Answers for Class 6Class 6 ScienceClass 6 EnglishClass 6 MathsClass 6 Social ScienceQuestion Answers for Class 5Class 5 ScienceClass 5 EnglishClass 5 MathsClass 5 Social ScienceQuestion Answers for Class 4Class 4 ScienceClass 4 EnglishClass 4 MathsA buffer solution consists of $N{H_3}$ and $N{H_4}Cl$, total concentration of buffering agent$ = 0.6M$ . If the $pOH$ of buffer solution is $5.05$ then calculate $[N{H_3}]$ solution. ($p{K_b}$ of $N{H_3} = 4.75$)A. $0.5$ B. $0.3$ C. $0.2$ D. $0.4$ AnswerVerified562.5k+ viewsHint: A buffer solution (more precisely, pH buffer or hydrogen ion buffer) is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. Its pH changes very little when a small amount of strong acid or base is added to it. Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety of chemical applications.Complete answer: In the above question, the buffer consists of ammonia and ammonium chloride and the concentration of this buffer solution is equal to $0.6M$ .Given, the $pOH$ of the buffer solution = $5.05$ $[Base] + [salt] = 0.6M$ Let the concentration of the salt be $x$ .Let the concentration of the base be $(0.6 - x)$.As per the Henderson-Hasselbalch equation,$pOH = p{K_b} + \log \left( {\dfrac{{[salt]}}{{[base]}}} \right)$ According to the question, the $p{K_b}$ of $N{H_3} = 4.75$ Thus, substituting the values in the above equation, we have:$ \Rightarrow 5.05 = 4.75 + \log \left( {\dfrac{x}{{0.6 - x}}} \right)$Thus, on solving, we have:$ \Rightarrow 0.3 = \log \dfrac{x}{{0.6 - x}}$ $ \Rightarrow anti\log (0.3) = \dfrac{x}{{0.6 - x}}$Also, $anti\log (0.3) = 2$ Thus, substituting the value in the above equation, we have:$ \Rightarrow \dfrac{x}{{0.6 - x}} = 2$ On solving, we get: $x = \dfrac{{1.2}}{3} = 0.4$ Thus, the concentration of the salt = $0.4M$ The concentration of the base = $0.6 - 0.4 = 0.2M$ Now, again applying the concentration of the salt in the Henderson-Hasselbalch equation, we can find the concentration of the base ($N{H_3}$ ) in the solution. Thus, the equation can be written as: \[ \Rightarrow 5.05 = 4.75 + \log \left( {\dfrac{{0.6}}{{[base]}}} \right)\]$ \Rightarrow 2 = \dfrac{{0.6}}{{[base]}}$ Thus, the concentration of $N{H_3}$ = $[N{H_3}] = 0.3M$ Thus option B is the correct answer.Note:The Henderson–Hasselbalch equation can be used to calculate the pH of a solution containing the acid and one of its salts, that is, of a buffer solution. With bases, if the value of equilibrium constant is known in the form of a base association constant, Kb the dissociation constant of the conjugate acid may be calculated from the given equation:$p{K_a} + p{K_b} = p{K_w}$ Recently Updated PagesMaster Class 12 Business Studies: Engaging Questions & Answers for SuccessMaster Class 12 Economics: Engaging Questions & Answers for SuccessMaster Class 12 English: Engaging Questions & Answers for SuccessMaster Class 12 Maths: Engaging Questions & Answers for SuccessMaster Class 12 Social Science: Engaging Questions & Answers for SuccessMaster Class 12 Chemistry: Engaging Questions & Answers for SuccessMaster Class 12 Business Studies: Engaging Questions & Answers for SuccessMaster Class 12 Economics: Engaging Questions & Answers for SuccessMaster Class 12 English: Engaging Questions & Answers for SuccessMaster Class 12 Maths: Engaging Questions & Answers for SuccessMaster Class 12 Social Science: Engaging Questions & Answers for SuccessMaster Class 12 Chemistry: Engaging Questions & Answers for Success

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