A 0.2g Sample Of Benzoic Acid C6H5COOH Is Titrated With ... - Vedantu

CoursesCourses for KidsFree study materialOffline CentresMoreStore

Talk to our experts

1800-120-456-456

Sign In

• Question Answer

• Class 11

• Chemistry

• A 02g sample of benzoic acid C...

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 0.2g sample of benzoic acid \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\] is titrated with a 0.120M \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\]solution. What volume of the \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\]solution is required to reach the equivalence point? Molar mass of $C_6H_5COOH$ = $122.1\, gmol^{-1}$A. 6.82 mlB. 13.6 ml C. 17.6 ml D. 35.2 ml AnswerVerified575.4k+ viewsHint: Write the balanced acid-base reaction between \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\] and \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\]. Calculate the moles of benzoic acid and then using the stoichiometric ratio, calculate the moles of \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\]. Finally using the moles and molar concentration of \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\] , calculate the volume of it. Formula Used: \[{\text{Moles = }}\dfrac{{{\text{mass}}}}{{{\text{Molar mass}}}}\]\[{\text{Molarity = }}\dfrac{{{\text{ moles }}}}{{{\text{ L of solution}}}}\]Complete step by step answer: The balanced acid-base reaction between \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\] and \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\] is as follows: \[{\text{2}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH (aq) + Ba(OH}}{{\text{)}}_{\text{2}}}{\text{(aq) }} \rightleftharpoons {\text{ Ba(}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COO}}{)_2}({\text{aq) + 2}}{{\text{H}}_{\text{2}}}{\text{O(l)}}\]Now, using the mass and molar mass of \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\] given to us calculate the moles of \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\]. \[{\text{Moles = }}\dfrac{{{\text{mass}}}}{{{\text{Molar mass}}}}\]Substitute 0.2g for the mass of \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\] and \[122.1{\text{ gm}}{{\text{ol}}^{{\text{ - 1}}}}\] for molar mass of \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\] and calculate the moles of \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\] as follows: \[{\text{Moles = }}\dfrac{{0.2{\text{ g}}}}{{122.1{\text{ gm}}{{\text{ol}}^{{\text{ - 1}}}}}} = 0.0{\text{0163 mol}}\]Now, using these moles of \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\] and balanced chemical reaction calculate the moles of \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\] at as follows: From the balanced reaction, we can say that 2 moles of \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\] reacts with 1 mole of \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\]. So, \[0.0{\text{0163 mol }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}} \times \dfrac{{1{\text{ mol Ba(OH}}{{\text{)}}_{\text{2}}}{\text{ }}}}{{2{\text{ mol }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH }}}} = 8.19 \times {10^{{\text{ - 4}}}}{\text{ mol Ba(OH}}{{\text{)}}_{\text{2}}}\]Now, we have moles of \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\] and also we have given molar concentration of \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\]. Hence, calculate the volume of \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\]require to reach the equivalence point as follows: \[{\text{Molarity = }}\dfrac{{{\text{ moles }}}}{{{\text{ L of solution}}}}\]Substitute \[8.19 \times {10^{{\text{ - 4}}}}{\text{ mol Ba(OH}}{{\text{)}}_{\text{2}}}\] and 0.120M \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\] in the molarity equation and calculate the volume of \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\]require to reach the equivalence point. \[{\text{Litres of Ba(OH}}{{\text{)}}_{\text{2}}}{\text{ solution}} = 0.00682{\text{ L}}\]Convert the volume of \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\] solution from L to ml. 1 L = 1000 ml \[0.00682{\text{ L}} \times \dfrac{{1000{\text{ ml}}}}{{1{\text{ L}}}} = 6.82{\text{ ml}}\]Thus, 6.82 ml of \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\] is required to reach the equivalence point. Hence, the correct option is (A) 6.82 mlNote: Acid is a proton donor species and the base is a proton acceptor species. It is very important to write the correct balance reaction as a mole calculation depends on the stoichiometric ratio. From the balanced reaction, we can say that at the equivalence point 1 mole of \[{\text{Ba(OH}}{{\text{)}}_{\text{2}}}\] reacts with 2 moles of\[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\].Recently Updated PagesMaster Class 11 Business Studies: Engaging Questions & Answers for SuccessMaster Class 11 Computer Science: Engaging Questions & Answers for SuccessMaster Class 11 Economics: Engaging Questions & Answers for SuccessMaster Class 11 Social Science: Engaging Questions & Answers for SuccessMaster Class 11 Chemistry: Engaging Questions & Answers for SuccessMaster Class 11 English: Engaging Questions & Answers for SuccessMaster Class 11 Business Studies: Engaging Questions & Answers for SuccessMaster Class 11 Computer Science: Engaging Questions & Answers for SuccessMaster Class 11 Economics: Engaging Questions & Answers for SuccessMaster Class 11 Social Science: Engaging Questions & Answers for SuccessMaster Class 11 Chemistry: Engaging Questions & Answers for SuccessMaster Class 11 English: Engaging Questions & Answers for Success

• 1
• 2

Trending doubtsOne Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSEDiscuss the various forms of bacteria class 11 biology CBSEDraw a diagram of a plant cell and label at least eight class 11 biology CBSEState the laws of reflection of lightExplain zero factorial class 11 maths CBSE10 examples of friction in our daily lifeOne Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSEDiscuss the various forms of bacteria class 11 biology CBSEDraw a diagram of a plant cell and label at least eight class 11 biology CBSEState the laws of reflection of lightExplain zero factorial class 11 maths CBSE10 examples of friction in our daily life

• 1
• 2

Repeaters Course for NEET 2022 - 23

NEET Repeater 2023 - Aakrosh 1 Year Course