Given that the titration of 20.0 mL of 0.200 M HNO3 with 0.100 M NaOH occurs via the following reaction: HNO3 + NaOH → Na* + NO3 + H₂O Find the pH at the equivalence point. (Show all your working.) Comment and explain the pH found in your answer Phenolphthalein, a colourless weak acid (K₁=6.31 x 10-¹0) which dissociates in water forming pink anions is widely considered a suitable indicator for this titration. Along with a suitable calculation, find the working/useful range of this indicator to explain its use.

Given that the titration of 20.0 mL of 0.200 M HNO3 with 0.100 M NaOH occurs via the following reaction: HNO3 + NaOH → Na* + NO3 + H₂O Find the pH at the equivalence point. (Show all your working.) Comment and explain the pH found in your answer Phenolphthalein, a colourless weak acid (K₁=6.31 x 10-¹0) which dissociates in water forming pink anions is widely considered a suitable indicator for this titration. Along with a suitable calculation, find the working/useful range of this indicator to explain its use.

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

Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution (a) before any NaOH is added? (b) at half-neutralization? (c) at the equivalence point? (d) when 0.10 mL less than the volume of NaOH to reach the equivalence point is added? (e) when 0.10 mL more than the volume of NaOH to reach the equivalence point is added? (f) Use your data to construct a plot similar to that shown in Figure 14.10 (pH versus volume NaOH added).
Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after the following additions of acid:(a) 0 mL(b) 15.00 mL(c) 29.00 mL(d) 29.90 mL(e) 30.00 mL(f ) 30.10 mL(g) 40.00 mL
calculate the pH during the titration of 30.00 mL of .1000 M KOH with .1000 M HBr solution after the following additions of acid: a. 0 mL b. 18.00 mL c. 30.00 mL d. 35.10 mL
Consider the titration of 25 mL of 0.15 M methylamine, CH3NH2, which has a Kb of 3.4 x 10-4 with 0.19 M HCl. Clearly calculate the initial pH.
In the titration of 25.0 mL of 0.15 M butanoic acid, CH3CH2CH2CO2H, with 0.20 M sodium hydroxide, what will be the pH at the equivalence point? The Ka for butanoic acid is 1.5 x 10-5. The correct answer should be reported to two places past the decimal.
Calculate the pH during the titration of 40.00 mL of 0.0250 M KOH with 0.0500 M HCl after addition of the following volumes of titrant reagent (HCl): a) 0.00 mL b) 10.00 mL c) 15.00 mL d) 18.00 mL e) 20.00 mL f) 25.00 mL
You are going to titrate 29.0 mL of 0.294 M cyanic acid (HCNO) with 0.15 M NaOH. What is the pH of the solution after 5.684 mL of the NaOH has been added? The Ka of cyanic acid at the temperature of the experiment is 2.3 x 10-4. Report your answer to the hundredths place and do not include units
Write the reaction between formic acid (Appendix B) and KOH. What is the equivalence volume (Ve) in the titration of 50.0 mL of 0.050 0 M formic acid with 0.050 0 M KOH? Calculate the pH at the points Vb 0.0, 10.0, 20.0, 25.0, 30.0, 40.0, 45.0, 48.0, 49.0, 49.5, 50.0, 50.5, 51.0, 52.0, 55.0, and 60.0 mL. Draw a graph of pH versus Vb. Without doing any calculations, what should the pH be at Vb5 12 Ve? Does your calculated result agree with the prediction?
Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH2)3N, with 0.1000 M HCl solution after the following additions of titrant. 10.00 mL: 21.00 mL: 28.00 mL:
In the quadruplicate titration of 0.2068 g; 0.1997g; 0.2245 g and 0.2137 g of sodium carbonate 36.31 mL were spent; 35.11mL; 39.00 mL and 37.54 mL of perchloric acid. a) Calculate the average molarity of the perchloric acid solution. b) Would there be statistical justification for rejecting the furthest molarity? Tabular Q = 0.829 for n = 4 and 95% confidence.
Consider the titration of 25 mL of 0.1 M HCHO2 (known as formic acid and has a Ka of 1.8 x 10-4) with 0.1 M of the strong base KOH. Clearly calculate the initial pH, the pH at half way to the equivalence point, the pH at the equivalence point, and the pH after the addition of infinite excess KOH.
Consider the titration of 25 mL of 0.15 M methylamine, CH3NH2, which has a Kb of 3.4 x 10-4 with 0.19 M HCl. Clearly calculate the pH at half way to the equivalence point and the pH at the equivalence point.