10-6. Consider the titration of 100.0 mL of 0.100 M NaOH with 1.00 M HBr. What is the equivalence volume? Find the pH at the following volumes of HBr and make a graph of pH versus Va: Va = 0, 1.00, 5.00, 9.00, 9.90, 10.00, 10.10, and 12.00 mL.
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- Consider the titration of 100.0 mL of 0.100 M NaOH with 1.00 M HBr. What is the equivalence volume? Find the pH at the following volumes of HBr and make a graph of pH versus Va: Va 0, 1.00, 5.00, 9.00, 9.90, 10.00, 10.10, and12.00 mL.Construct a curve for the titration of 50.0 mL of 0.100 M hydrazoic acid, HN3 (Ka = 2.2 x 10-5) with 0.200 M solution of KOH. Calculate the pH after the addition of 0.00,12.50, 20.00, 24.00, 25.00, 26.00, 37.50, 45, and 50.00 mL of titrant. Identify the specie/species present at the following regions of the curve:a. Start of the titrationb. Before equivalence pointc. At equivalence pointd. Past the equivalence point.Construct a curve for the titration of 20.00 mL 0.500 N succinic acid (HO2CCH2CH2CO2H) by 0.1000 N NaOH solution. Use MS Excel for convenience. Titrant volumes: 0.00 mL, 5.00 mL, 10.00 mL, 15.00 mL, 20.00 mL, 25.00 mL, and 30.00 mL. (CH2)2(CO2H)2⇄(CH2)2(CO2H)(CO2)^- +H^+ Ka1= 6.21 x 10^-5; pKa1 =4.207 (CH2)2(CO2H)(CO2)^-⇄(CH2)2(CO2)2^2-+ H^+ Ka2= 2.31 x 10^-6; pKa2 =5.636
- If a 0.3 M solution of NaOH was used to titrate 100 mL of 0.3 M solution of CH3COOH (Ka =1.8 x 10 -5) and the following volumes of NaOH were recorded: 102, 97, 99, 98, 101, 106 mL. The pH Value is………………when added 99.0 mL of 0.3 M NaOH The pH of the solution at the equivalence point more than 7 Because................................... Calculate the 99% confidence limits of the mean? And use them to decide whether there is any evidence of systematic error? [ hints: For N=6 and the 99 % confidence interval, the value of t is 4.03].Consider the titration of 300.0 mL of 0.500 M NH3 (Kb = 1.8 x 10-5) with 0.500 M HNO3. What is the pH of the solution when 248 mL of 0.500 M HNO3 has been added to the base? **Report your answer to two decimal places. DO NOT worry about significant figures.**Consider the titration of 50.00 mL of 0.100 M CH3CO2H with 0.200 M KOH, calculate the pH after adding 30.00 mL of KOH. Ka for CH3CO2H = 1.8 x 10-5. Group of answer choices 12.10 1.90 11.50 7.00
- What is the equivalence volume in the titration of 50.00 mL of 0.010 0 M NaOH with 0.100 M HCl? Calculate the pH at the following points: Va 0.00, 1.00, 2.00, 3.00, 4.00, 4.50, 4.90, 4.99, 5.00, 5.01, 5.10, 5.50, 6.00, 8.00, and 10.00 mL. Make a graph of pH versus Va.Construct a curve for the titration of 50.0 mL of 0.100 M hydrazoic acid, HN3 (Ka = 2.2 x 10-5) with 0.200 M solution of KOH. Calculate the pH after the addition of 0.00, 12.50, 20.00, 24.00, 25.00, 26.00, 37.50, 45, and 50.00 mL of titrant. Identify the specie/species present at the following regions of the curve:1. Start of the titration2. Before equivalence point3. At equivalence point4. Past the equivalence point.A solution made by mixing 105.00 mL of 1.050 M CH3COOH with 35.00 mL of 0.750 M NaOH. For CH3COOH, Ka = 1.8×10-5. find the pH.
- 50 mL of benzoic acid (HC7H5O2) 0.1 mol L-1 are titrated with NaOH 0.1 mol L-1. Calculate the pH after addition of 0, 20, 50 and 100 mL of NaOH (Ka = 6.31 x 10-5). Generate the titration curve and choose the best indicator for this procedure.Note: possible answers are 2.60; 3.60; 4.20 and 8.45. Although there are the answers, show the calculations and generate the titration curve.1Upon the addition of 30.00 mL of 0.350 M HCl to a 80.00-mL of the weak base amine (Kb = 4.95 x 10-10) whose concentration is 0.530M, the resulting pH is then equal to 8.80. True or False? 2.When 100 mL of a 1.0M strong acid is 50% titrated with 1.0M strong base, the volume of the base should be added is 50mL and this corresponds the equivalence point. True or False? 3. The pH of a buffer can be calculated using the formula: 14-pOH = pKb + log of [proton acceptor]/[acid component]. True or False?Find the pH during the titration of 20.00 mL of 0.2850 M benzoic acid, C6H5COOH (Ka = 6.3 10-5), with 0.2850 M NaOH solution after the following additions of titrant. (a) 0 mL(b) 10.00 mL(c) 15.00 mL(d) 20.00 mL(e) 25.00 mL