Question 2 15 P Weak Acid Titration Calculations. A 25.0 mL aliquot of 0.50 M propanoic...
. 20.0 mL of 0.100 M lactic acid solution is titrated with 0.100 M NaOH solution. Calculate the pH of the contents of the Erlenmeyer flask at each of the following points during the titration. (a) When 0.00 mL of NaOH have been added. (2) (b) After 5.00 mL of NaOH have been added. (3) (c) After 20.0 mL of NaOH have been added. (3) (d) After 10.0 mL of NaOH have been added. (1) (e) After 25.0 mL of...
Exactly 25.00 ml of a 0.0685 M pentanoic acid solution,(C4H8O2), is placed in a flask and titrated with standardized 0.050 M NaOH. (the pKa of this acid is 4.84). A. Determine the endpoint of the titration B. Determine the pH of the solution after the following amounts of NaOH are added to the acid solution: a.) Initially, 0.00 ml NaOH. b.) 5.00 ml c.) 15.00 ml d.) 25.00 ml e.) 34.00 ml...
a 25.0 ml aliquot of vinegar was diluted to 250 ml in volumetric flask. Titration of 50.0 ml aliquots of diluted solution required an average of 34.88 ml of 0.09600 M NaOH. Express the acidity of the vinegar in terms of percentage (w/v) of acetic acid.
16. Exactly 25.00 ml of 0.0685 M HClO4 solution is placed in a flask and titrated with standardized 0.050M NaOH. A. Determine the endpoint of the titration B. Determine the pH of the solution after the following amounts of NaOH are added to the acid solution: a.) Initially, 0.00 ml NaOH. b.) 5.00 ml c.) 15.00 ml d.) 25.00 ml e.) 34.00 ml f.) At the endpoint of the titration g.)...
A 20.00 ml aliquot of a 1.00 M solution of a weak acid (Ka = 1.50E-04 at 25ºC) is titrated with 0.500 M NaOH. Determine the pH at the following volumes of NaOH. Volume NaOH = 30.00mL , pH? Volume NaOH = 40.00mL , pH? Volume NaOH = 47.50mL , pH?
A 50.00 ml aliquot of 0.1000 M NaOH is titrated with 0.1000 M HCL. Calculate the pH of the solution after the addition of 0.00, 10.00, 25.00, 40.00, 45.00, 49.00, 50.00, 51.00, 55.00, and 60.00 ml of acid and prepare a titration curve from the data.
Acid/Base titrations 10. 25.0 mL of 0.100 M H2A (a weak diprotic acid) is titrated with 0.200 M NaOH. What is the pH of the solution when 0.00 mL, 10.0 mL, 12.5 mL, 20.0 mL, 25.0 mL, and 40.0 mL E.S RaWeMA 5.83 x 10 8) have been added? (Ka1= 2.46 x 10, Ka2 ANSWER: 0 mL = 2.315, 10.0 mL= 4.211 (or 4.213), 12.5 mL = 5.422, 20.0 mL 7.410, 25.0 mL = 9.966, 40.0 mL = 12.664 11....
A 120.0-mL aliquot of 0.114 M weak base B (pKb = 4.58) was titrated with 1.14 M HClO4. Find the pH at the following volumes of acid added: Va = 0.00, 1.30, 6.00, 11.00, 11.90, 12.00, 12.10, and 17.00 mL. (Assume Kw = 1.01 ✕ 10−14.)
Part A: Calculating a Theoretical Titration Curve (Weak Acid - Strong Base) Consider the titration of 50.00 mL of 0.05 M acetic acid with 0.1 M NaOH. Calculate the pH of the resulting solution at the following points during the titration (given as volume of NaOH added). Volume NaOH pH of analyte 0.00 15.00 20.00 24.00 24.50 mL at equivalence point 40.00
A 100.0 mL aliquot of 0.100 M monoprotic acid H2A (pK1 = 4.00) was titrated with 1.00 M NaOH. Find the pH at the following volumes of base added: Vb = 0, 5, 9, 10 and 11 mL. Sketch the titration curve.