Please answer the question all information is provided Data and Calculations1. Measurement of pH and titration of acetic acid solutionConcentration of standardized NaOH titrant 0. 1029mol/LMass concentration of acetic acid2.40g/LTrial 1Trial 2Trial 3Measured pH of the acetic acid solution3.103.123.1230.0830.0430.05Mass of acetic acid solution taken for titrationgggInitial buret reading of NaOH titrant2.3014.025.737.311.61.191.190.0396 mol/L60.61 g/molmLmLmL14.0I1.71.20Final buret reading of NaOH titrant25.711.7mLmLmLNet volume of NaOHFinnl-initolmLmLmLMillimoles of NaOH to end point of titrationmmol1.20mmolmmolMillimoles of acetic acid in sample1.20 mmol|.20mmolmmol0.040 mol/L60.16 8/molMolar concentration of acetic acid solution0.040mol/LCalculated molar mass of acetic acid(0.08 8/molCalculation of the ionization constant for acetic acid from the measured pH of the acetic acidsamples. (Average the pH and molar concentration values for the samples of acetic acid you titrated; use theaverage values in the calculation of the ionization constant.)60.28Average molar concentration of acetic acid for the samples you titrated:mol/L3.1Average pHCalculate the corresponding [H3O*], compute the concentration of A¯ and HA, and calculate the dissociationconstant, K,.M H3O+M AМ НА[H;O*][A]KaНА)pk, =

Calculation of the ionization constant for acetic acid from the measured pH of the acetic acid samples. (Average the pH and molar concentration values for the samples of acetic acid you titrated; use the average values in the calculation of the ionization constant.) 60.28 Average molar concentration of acetic acid for the samples you titrated: mol/L 3.11 Average pH Calculate the corresponding [H3O*], compute the concentration of A and HA, and calculate the dissociation constant, Ką. M H30* МА- М НА [H;O*][A] K. НА pK,

Calculation of the ionization constant for acetic acid from the measured pH of the acetic acid samples. (Average the pH and molar concentration values for the samples of acetic acid you titrated; use the average values in the calculation of the ionization constant.) 60.28 Average molar concentration of acetic acid for the samples you titrated: mol/L 3.11 Average pH Calculate the corresponding [H3O], compute the concentration of A and HA, and calculate the dissociation constant, Ką. M H30 МА- М НА [H;O*][A] K. НА pK,

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter15: Acid–base Equilibria

Section: Chapter Questions

Problem 107AP

See similar textbooks

Similar questions

In an acid-base titration, 21.16 mL of an NaOH solution are needed to neutralize 20.04 mL of a 0.0997 M HCl solution. To find the molarity of the NaOH solution, we can use the following procedure: First note the value of MH in the HCl solution. ____________M Find MOH- in the NaOH solution. Use Eq.3. ____________M Obtain MNaOH from MOH. ____________M
please stop rejecting. we pay for this service, so we deserve getting our answer! I do not want to have to put a claim in and report Use acid-base titration to determine theconcentration of:– A strong acid: HCl 0.100 M NaOH (standardized)•0.100 M HCl0.100 M CH3COOH Phenolphthalein is used for endpoint determination– Changes from colourless to pink as a solution becomes more basic• Only need 2-3 drops per titration• pH range: 8.3-10
The titration of 5.00 mL of 0.4949 M NH3 with 0.2250 M HNO3. Kb = 1.8×10–5. Fill out chart and list possible indicator. Volume HNO3 (mL) Calculated pH 1 initial 0.00 2 ¼ way to eq. point 3 ½ way to eq. point 4 ¾ way to eq. point 5 0.5 mL before eq. point 6 equivalence point 7 0.5 mL after eq. point 8 well past eq. point 15.00
Titration of (4.890x10^-1) grams of monoprotic weak acid (dissolved in KCl) required (2.44000x10^1) mL of (2.12000x10^-1) molar NaOH solution to reach equivalence. Based on this information, how many moles of weak acid are present in this solution? Format your answer with correct number of sig figs. Use scientific notation.
A sample may contain NaOH, sodium carbonate and sodium bicarbonate, alone or in compatible combination. Titration of a 50.00 mL aliquot of the sample to a phenolphthalein end point required 8.50 mL of 0.0905 M HCl. A second 50.00 mL aliquot required 22.40 mL of the HCl when titrated to a methyl red end point. Calculate the molar concentration of the base component/s
6. What do we call reagents of known concentration that is used to carry out a volumetric titration?A. standard titrateB. standard analyteC. standard titrantD. standard reagent7. What has the analyte reached when the amount of added titrant is chemically equivalent to the amount of the analyte in the sample?A. equivalence pointB. end-pointC. titration pointD. Dilution point8. In an acid-base titration, what indicator does an analytical chemist add to indicate chemical equivalence?A. bromophenol greenB. phenolphthaleinC. iodineD.methyl orange9. These are factors in the environment that causes exposure to the emission of electromagnetic waves.A. BiologicalB. ChemicalC. PhysicalD. Radiation10. What concentration unit measures the number of chemical equivalents of reagent in the same volume?A. MolalityB. FormalityC. MolarityD. Normality
Titration Titrant Reductant/ Oxidant Analyte Standardizing Indicator Reagent COD BOD INDIRECT TITRATIONKindly identify the blank sections appropriately.
A 250.0mg sample of an organic monoprotic weak acid was dissolved in an appropriate solvent and titrated with 0.091M NAOH , requiring 29.5ml to reach end point. Determine equivalent weight
What volume (mL) of 0.2173M NaOH is required to titrate 50.00mL of 0.2980M HCI ?In the titration of 50.00mL of 0.2980M HCl. 71.93mL of 0.2173M NaOH was required to reach the phenolphthalien endpoint. Calculate the percent error between the predicted amount and the experimental amount. What does this tell you about the concentration of the NaOH? This is based on your answers to questions 1 and 2. Question 3 options: The concentration of the NaOH is 0.2173M. The concentration of the NaOH is stronger that 0.2173M. Nothing new The concentration of the NaOH is weaker than 0.2173M. Please answer fast i give upvote.
Consider the Heat of Solution Practical that was conducted. Temperature (oC) Amount of NaOH ( 0.015M) used to reach end point of titration (ml) 40 42 35 35 30 26 25 23 20 16 Use the data provided above in order to determine the heat of solution. Include any necessary graphs.
The data below is titration of Na2CO3 with HCl which is a diprotic titration curve. From the endpoint determinations, calculate the concentration of the carbonate solution if the [HCl] was 0.1036 M. use 10.00 mL of carbonate solution for the calculations PH Burett vol added 11.30 0.00 0.00 10.80 0.98 0.98 10.51 1.95 0.97 10.28 3.00 2.02 10.10 3.99 3.01 9.92 4.99 4.01 9.74 5.95 4.97 9.54 6.97 5.99 9.28 7.95 6.97 8.86 8.97 7.99 8.70 9.18 8.2 8.52 9.35 8.37 8.39 9.49 8.51 8.26 9.60 8.62 8.09 9.71 8.73 7.97 9.82 8.84 7.90 9.90 8.92 7.93 10.21 9.23 7.87 10.32 9.34 7.84 10.42 9.44 7.75 10.53 9.55 7.65 10.63 9.65 7.57 10.75 9.77 7.53 10.85 9.87 7.49 10.96 9.98 7.05 11.95 10.97 6.76 12.99 12.01 6.56 13.98 13 6.39 14.93 13.95 6.24 15.98 15 6.07 16.98 16 6.21 17.05 16.07 6.24 17.15 16.17 6.25 17.25 16.27 6.30 17.32 16.34 6.37 17.40 16.42 6.41 17.50 16.52 6.46 17.58 16.6 6.47 17.63 16.65 6.45 17.72 16.74 6.44…