Standardization of NaOHTrial 1Trial 2Trial 3Trial 4Trial 5D. 785O.l690g0. 8239Mass of KHPFinal buret16.30mL: 14.05ML 17.10MLreadingInitial buret0,00mL 0.0vmLO.comLreadingVolume of NaOHCalculate the molarity of the NaOH for each trial:Trial 1Trial 2Trial 3Trial 4Trial 5CalculatedMolarityAverage MolarityM

The mass of KHP used in trial 1, trial 2 and trial 3 is 0.785 g, 0.690 g and 0.823 g, respectively. The final reading of burette in trial 1, trail 2 and trial 3 is 16.30 mL, 14.05 mL and 17.10 mL, respectively.The reaction that takes place between KHP and NaOH is expressed as, NaOH+C8H5KO4→C8H4KO4Na+H2O At equivalence point, nKHP=nNaOH......(1) The volume of NaOH used during trial 1 is calculated as, VNaOH=16.30 mL-0.00 mL=16.30 mL=0.0163 L The volume of NaOH used during trial 2 is calculated as, VNaOH=14.05 mL-0.00 mL=14.05 mL=0.01405 L The volume of NaOH used during trial 3 is calculated as, VNaOH=17.10 mL-0.00 mL=17.10 mL=0.0171 LThe number of moles of KHP is calculated by the formula, nKHP=mKHPMKHP......(2) Where, mKHP is the mass of KHP. MKHP is the molar mass of KHP (204.22 g/mol). For trial 1 substitute the mass and molar mass of KHP in equation (2). nKHP=0.785 g204.22 g/mol=0.0038438 mol For trial 2 substitute the mass and molar mass of KHP in equation (2). nKHP=0.690 g204.22 g/mol=0.0033787 mol For trial 3 substitute the mass and molar mass of KHP in equation (2). nKHP=0.823 g204.22 g/mol=0.0040299 molAccording to equation (1), the number of moles of NaOH for trial 1, trial 2 and trial 3 is 0.0038438 mol, 0.0033787 mol and 0.0040299 mol, respectively. The molarity of NaOH is calculated by the formula, M=nNaOHVNaOH......(3) For trial 1 substitute the number of moles and volume of NaOH in equation (3). M1=0.0038438 mol0.0163 L=0.236 M For trial 2 substitute the number of moles and volume of NaOH in equation (3). M2=0.0033787 mol0.01405 L=0.240 M For trial 3 substitute the number of moles and volume of NaOH in equation (3). M3=0.0040299 mol0.0171 L=0.236 MThe average molarity of NaOH is calculated as, M=M1+M2+M33=0.236 M+0.240 M+0.236 M3=0.237 MAnswer: The molarity of NaOH for trial 1, trial 2 and trial 3 is 0.236 M, 0.240 M and 0.236 M, respectively. The average molarity of NaOH is 0.237 M.

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Standardization of NaOH Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 0.785 0. l690g 0.8239 Mass of KHP Final buret 16.30mL 14.05mL 17.10mL reading Initial buret 0.00mL 0.00mL 0.UDmL reading Volume of NaOH Calculate the molarity of the NaOH for each trial: Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Calculated Molarity Average Molarity M

Standardization of NaOH Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 0.785 0. l690g 0.8239 Mass of KHP Final buret 16.30mL 14.05mL 17.10mL reading Initial buret 0.00mL 0.00mL 0.UDmL reading Volume of NaOH Calculate the molarity of the NaOH for each trial: Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Calculated Molarity Average Molarity M

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

Standardization of NaOH
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
D. 785
O.l690g
0. 8239
Mass of KHP
Final buret
16.30mL: 14.05ML 17.10ML
reading
Initial buret
0,00mL 0.0vmL
O.comL
reading
Volume of NaOH
Calculate the molarity of the NaOH for each trial:
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
Calculated
Molarity
Average Molarity
M

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