Chem 1320 1st Edition Lecture 9Outline of Last Lecture I. Key TermsII. HydrationIII. Chemistry in SolutionA. QualitativeB. QuantitativeOutline of Current Lecture I. Word ProblemCurrent LectureI. Word ProblemIf there is a 2.324g sample of an unknown acid, and H2X is dissolved in water. The sample requires 68.83mL of a 0.7500M NaOH solution for a complete reaction. What is the molar mass of the acid?Molar mass = mass/molH2X (aq) + 2NaOH (aq) Na2X (aq) + 2H2O (l) (BALANCED)Net Ionic Equation: 2H+ (aq) + X2- (aq) + 2Na+ (aq) + 2OH- (aq) 2Na+(aq) + X2-(aq) +2H20 (l)Cancel all the identical terms on each side: 2H+ (aq) + 2OH- (aq) 2H20 (l)Re-Balance Equation: H+ (aq) + OH- (aq) H2O (l)M x V = mol/L x L(68.85mL) (1L) (0.7500mol) (1000 mL) (1L) = 5.162 x 10-2 mol OH-These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.(0.05)(2 mol H2X/ 1 mol NaOH) 0.0258 mol H2XMolar mass = mass/mol 2.324g/0.025g =
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