1)A 30.0-mL sample of 0.165 Mpropanoic acid (CH3CH2COOH, Ka = 1.3 × 10−5) is titrated with 0.300 MKOH. d)Choose a volume (in mL) of potassium hydroxide added that is 2-9 mL of titrant added before the equivalence point and calculate the pH at this volume. VKOH added = _________ mL e)Calculate the pH at the equivalence point. f)Choose a volume (in mL) of potassium hydroxide added that is 2-9 mL of titrant added after the equivalence point and calculate the pH at this volume. VKOH added = _________ mL

1)A 30.0-mL sample of 0.165 Mpropanoic acid (CH3CH2COOH, Ka = 1.3 × 10−5) is titrated with 0.300 MKOH. d)Choose a volume (in mL) of potassium hydroxide added that is 2-9 mL of titrant added before the equivalence point and calculate the pH at this volume. VKOH added = _ mL e)Calculate the pH at the equivalence point. f)Choose a volume (in mL) of potassium hydroxide added that is 2-9 mL of titrant added after the equivalence point and calculate the pH at this volume. VKOH added = _ mL

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter14: Equilibria In Acid-base Solutions

Section: Chapter Questions

Problem 74QAP: Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution (a)...

See similar textbooks

Similar questions

The following results were obtained in the titration of an unknown monoprotic acid (Use data from table). Students identified the acid as benzoic. Copy VNaOH (no first raw), then pH(all values) into your Lab Report template. Use the data you obtained to fill in the table below. Concentration of NaOH used, M 1.164 x 10-1 Mass of unknown aid used, g 0.3840 1) Find: a) Equivalence point, mL. b) Half equivalence point, mL. c) pKa d) Ka e) pKa from literature f) Percent Error g) Number of moles of NaOH, mol. h) Number of moles of acid titrated, mol. i) Molar mass of acid g/mol. j) Molar mass of acid, g/mol from the literature k) Percent error
Consider the titration of 25.00mL of 0.200M methylamine (CH3NH2). The titrant is 0.120M HCl. (Calculate each below using ICE and ICF tables.) Note you should NOT use M1V1=M2V2 (a) the initial pH of base? (b) the pH at 5.00mL added? (c) the pH at 1/2 the HCl needed to reach the equivalence point. (d) the volume of added acid required to reach the equivalence point? (e) the pH at equivalence point.
Be sure to answer all parts. Find the pH and the volume (mL) of 0.487 M HNO3 needed to reach the equivalence point in the titration of 2.65 L of 0.0750 M pyridine (C5H5N, Kb = 1.7 × 10−9). Volume = mL HNO3 pH =
Determine the pH at the point in the titration of 40.0 mL of 0.200 M HC₄H₇O₂ with 0.100 M Sr(OH)₂ after 100.0 mL of the strong base has been added. The value of Ka for HC₄H₇O₂ is 1.5 × 10⁻⁵. Is my ICE table correct? if not can you help me correct it by creating an ICE TABLE? and determine the pH base on the result of the acid-base reaction? thank you!
What volume of 0.00500 M weak acid is needed to titrate 40.00 mL of 0.0100 M barium hydroxide to the equivalence point? Group of answer choices 160. mL 10.0 mL 80.0 mL 40.0 mL The pKb for barium hydroxide must be provided in order to solve this.
Two 21.0 mLmL samples, one 0.100 molL−1HClmolL−1HCl and the other 0.100 molL−1HFmolL−1HF, were titrated with 0.200 molL−1KOHmolL−1KOH. a)What is the volume of added base at the equivalence point for HClHCl? b) -What is the volume of added base at the equivalence point for HFHF? c) Predict whether the pHpH at the equivalence point for each titration will be acidic, basic, or neutral. Predict whether the at the equivalence point for each titration will be acidic, basic, or neutral. neutral for HClHCl, and acidic for HFHF neutral for HClHCl, and basic for HFHF neutral for HFHF, and basic for HClHCl neutral for HFHF, and acidic for HClHCl neutral for both d) Predict which titration c urve will have the lowest initial pHpH. Predict which titration curve will have the lowest initial . HFHF curve HClHCl curve e)Choose a rough sketch of each titration curve.
he titration of 20.0-mL sample of 0.105 M Succinic Acid (Ka= 6.16 X10^-5) with 0.125 M NaOH. The initial pH with 0 mL of NaOH added? The volume of added base required to reach the equivalence point? The pH at 11 mL of added base? The pH at one-half equivalence point? The pH at equivalence point? The pH after adding 21.50 mL of base beyond the equivalence point?
A 37.0 mL sample of 0.106 M HNO2 is titrated with 0.220 M KOH. (Kafor HNO2 is 4.57\times 10-4.) Find PH
A student was given an unknown solution monoprotic acid to analyze. The student decided to use potentiometric titration to analyze the pKa of the unknown acid. The student took a 50mL sample and used a standardized 0.155 M NaOH titrant to generate the data below: Vtit, mL pH Vtit, mL pH 2.0 3.42 8.5 4.27 3.0 3.53 9.5 4.66 4.0 3.61 10.5 7.71 5.0 3.74 11.5 9.11 6.0 3.87 12.5 9.95 7.0 4.05 13.5 10.36 What is the concentration of the unknown acid?
A student was given an unknown solution monoprotic acid to analyze. The student decided to use potentiometric titration to analyze the pKa of the unknown acid. The student took a 50mL sample and used a standardized 0.155 M NaOH titrant to generate the data below: Vtit, mL pH Vtit, mL pH 2.0 3.42 8.5 4.27 3.0 3.53 9.5 4.66 4.0 3.61 10.5 7.71 5.0 3.74 11.5 9.11 6.0 3.87 12.5 9.95 7.0 4.05 13.5 10.36 The equivalence volume using the second derivative plot is 10.2 mL. What is the concentration of the unknown acid in M?
A 47.9 ± 0.5 mL solution of unknown concentration of HNO3 is titrated with a 0.100 ±0.005 M standard solution of Sr(OH)2. If 13.3 mL of Sr(OH)2 is added to reach the equivalence point, what is the concentration of HNO3 in the unknown? Show all work and sig. figs. do matter.
If I titrate 150 ml of 1.10 M CH3COOH (Ka= 1.8 X 10-5) with 0.55 M Hydrobromic acid calculate the following: please answer all parts Initial pH of solution. pH after adding 25.0 ml of titrant. Amount of titrant needed to reach the equivalence point. pH at the equivalence point. pH after adding amount to reach the equivalence point + 5 ml of titrant