60.0 mL of 0.625 M HNO, is titrated by 0.200 M KOH.Calculate the pH of the acid solution before any titrant is added.pH =Calculate the pH after 141.8 mL of 0.200 M KOH is added to 60.0 mL of 0.625 M HNO3.pH=Calculate the pH after 187.5 mL of 0.200 M KOH is added to 60.0 mL of 0.625 M HNOpH=Calculate the pH after 318 mL of 0.200 M KOH is added to 60.0 mL of 0.625 M HNO3.pH= At which point in the titration is the number of moles of analyte and titrant the same?HdDA. O Point AB. Point BC. O Point CD. Point DABCVolume titrant added (mL)

First, let's write the chemical equation for the titration:HNO₃ + KOH → H₂O + KNO₃This equation…

Answered: 60.0 mL of 0.625 M HNO, is titrated by…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter14: Principles Of Neutralization Titrations

Section: Chapter Questions

Problem 14.17QAP

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Another way to treat data from a pH titration is to graph the absolute value of the change in pH per change in milliliters added versus milliliters added (pH/mL versus mL added). Make this graph using your results from Exercise 61. What advantage might this method have over the traditional method for treating titration data?
You are given the following acidbase titration data, where each point on the graph represents the pH after adding a given volume of titrant (the substance being added during the titration). a What substance is being titrated, a strong acid, strong base, weak acid, or weak base? b What is the pH at the equivalence point of the tiration? c What indicator might you use to perform this titration? Explain.
Follow the directions of Question 64. Consider two beakers: Beaker A has a weak acid(K a=1105). Beaker B has HCI. The volume and molarity of each acid in the beakers are the same. Both acids are to be titrated with a 0.1 M solution of NaOH. (a) Before titration starts (at zero time), the pH of the solution in Beaker A is the pH of the solution in Beaker B. (b) At half-neutralization (halfway to the equivalence point), the pH of the solution in Beaker A the pH of the solution in Beaker B. (c) When each solution has reached its equivalence point, the pH of the solution in Beaker A the pH of the solution in Beaker B. (d) At the equivalence point, the volume of NaOH used to titrate HCI in Beaker B the volume of NaOH used to titrate the weak acid in Beaker A.
Two samples of 1.00 M HCl of equivalent volumes are prepared. One sample is titrated to the equivalence point with a 1.00 M solution of sodium hydroxide, while the other sample is titrated to the equivalence point with a 1.00 M solution of calcium hydroxide. a Compare the volumes of sodium hydroxide and calcium hydroxide required to reach the equivalence point for each titration. b Determine the pH of each solution halfway to the equivalence point. c Determine the pH of each solution at the equivalence point.
A quantity of 0.15 M hydrochloric acid is added to a solution containing 0.10 mol of sodium acetate. Some of the sodium acetate is converted to acetic acid, resulting in a final volume of 650 mL of solution. The pH of the final solution is 4.56. a What is the molar concentration of the acetic acid? b How many milliliters of hydrochloric acid were added to the original solution? c What was the original concentration of the sodium acetate?
A quantity of 0.25 M sodium hydroxide is added to a solution containing 0.15 mol of acetic acid. The final volume of the solution is 375 mL and the pH of this solution is 4.45. a What is the molar concentration of the sodium acetate? b How many milliliters of sodium hydroxide were added to the original solution? c What was the original concentration of the acetic acid?
Phenol, C6H5OH, is a weak organic acid. Suppose 0.515 g of the compound is dissolved in enough water to make 125 mL of solution. The resulting solution is titrated with 0.123 M NaOH. C6H5OH(aq) + OH(aq) C6H5O(aq) + H2O() (a) What is the pH of the original solution of phenol? (b) What are the concentrations of all of the following ions at the equivalence point: Na+, H3O+, OH, and C6H5O? (c) What is the pH of the solution at the equivalence point?
A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After 23.75 mL of the weak acid solution has been added to 50.0 mL of the 0.100 M NaOH solution, the pH of the resulting solution is 10.50. Calculate the original concentration of the solution of weak acid.
A solution with a pH of 9.22 is prepared by adding water to 0.413 mol of KX to make 2.00 L of solution. What is the pH of the solution after 0.368 mol of HX is added?
Sulfanilic acid (NH2C6H4SO3H) is used in manufacturing dyes. It ionizes in water according to the equilibrium equation NH2C6H4SO3H(aq)+H2O(l)NH2C6H4SO3(aq)+H3O+(aq)Ka=5.9104 A buffer is prepared by dissolving 0.20 mol of sulfanilicacid and 0.13 mol of sodium sulfanilate (NaNH2C6H4SO3) in water and diluting to 1.00 L. Compute the pH of the solution. Suppose 0.040 mol of HCl is added to the buffer.Calculate the pH of the solution that results.
A solution made up of 1.0 M NH3 and 0.50 M (NH4)2SO4 has a pH of 9.26. a Write the net ionic equation that represents the reaction of this solution with a strong acid. b Write the net ionic equation that represents the reaction of this solution with a strong base. c To 100. mL of this solution, 10.0 mL of 1.00 M HCl is added. How many moles of NH3 and NH4+ are present in the reaction system before and after the addition of the HCl? What is the pH of the resulting solution? d Why did the pH change only slightly upon the addition of HCl?
Use Le Chátelier's principle to explain what happens to the equilibrium H2O(l)H+(aq)+OH(aq) when a few drops of HCl are added to pure water.

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