Assume a titration with 0.100 M NaOH titrant and 25.00 mL of a 0.0800 M CH3COOH analyte. How many mL of NaOH is required to reach the equivalence point?
Assume a titration with 0.100 M NaOH titrant and 25.00 mL of a 0.0800 M CH3COOH analyte. What will the initial pH of the analyte be if 0.00 mL of NaOH is added?
Assume a titration with 0.100 M NaOH titrant and 25.00 mL of a 0.0800 M CH3COOH...
Titration of 25.00 mL of 0.100 M HCl with 0.100 M NaOH (strong acid, strong base): Answer the following questions: 4. Calculate the initial pH 5 Why is pH = 7 at the equivalence point? 6Why does the pH rise slowly at first, very rapidly near the equivalence point, and slowly after the equivalence point? 7. Why does it require 25.00 mL of NaOH to reach the equivalence point?
Consider the titration of 25.00 mL of 0.200 M methyl amine (CH3NH2). The titrant is 0.120 M HCl. Calculate each of the following: a. the volume of added acid required to reach the equivalence point. b. the pH at ½ the HCl needed to reach the equivalence point c. the pH at the equivalence point.
The next 9 questions are related to the titration of 25.00 mL of a 0.1000 M acetic acid solution with 0.1000 M KOH. 1.What is the initial pH of the analyte solution? 2.What volume of KOH is required to reach the equivalence point of the titration (in mL)? 3.How many mmol of the salt are present at the equivalence point? (ANALYTICAL AMOUNT, NOT EQUILIBRIUM AMOUNT) 4.What is the volume of the solution at the equivalence point (in mL)? 5.What is...
1. Consider the titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH solution. What volume of NaOH is required to reach the equivalence point in the titration? a. 25.0 mL b. 50.0 mL c. 1.00 × 10^2 mL d. 1.50 × 10^2 mL 2. Consider the following acid–base titrations: I) 50 mL of 0.1 M HCl is titrated with 0.2 M KOH. II) 50 mL of 0.1 M CH3COOH is titrated with 0.2 M KOH. Which statement...
The following table shows the data for the titration of 25.00 mL of 0.200 M CH3COOH (pKa= 4.75) with 0.200 M NaOH (sodium hydroxide). Volume of NaOH added (mL) pH of the solution (measured) Mols of CH3COO-formed Mols CH3COOH remained pH of the solution (calculated) 0.00 2.72 0 0.005 2.70 10.00 4.57 0.002 0.003 4.57 12.50 4.7 0.0025 0.0025 4.75 24.90 7.14 0.0049 2 x 10-5 7.14 24.99 8.14 0.00 2 x 10-6 8.15 25.00 8.88 0.005 0 8.88 25.01...
Consider the titration of 25.00 mL of 0.08364 M pyridine with 0.1067 M HCl (a) What volume of the titrant must be added to reach the equivalence point? (b) Find the pH when 4.63 mL of the titrant has been added.
You have 25.00 mL of a 0.100 M aqueous solution of the weak base CH3NH2 (Kb = 5.00 x 10-4). This solution will be titrated with 0.100 M HCl. (a) How many mL of acid must be added to reach the equivalence point? (b) What is the pH of the solution before any acid is added? (c) What is the pH of the solution after 5.00 mL of acid has been added? (d) What is the pH of the solution...
E. Titration calculations 1. A 25.0-ml sample of 0.100 M HCl is titrated with 0.125 M NaOH. How many milliliters of the titrant will be need to reach the equivalence point? 2. A 25.0-ml sample of 0.100 M Ba(OH)2 is titrated with 0.125 M HCI. How many milliliters of the titrant will be need to reach the equivalence point? 3. For the following titrations, determine if the equivalence points will be acidic, basic, or neutral i. NH3 titrated with HCI...
Consider the titration of the titration of 50.0 mL of 0.100 M acetic acid (HC2H2O2) with 0.100 M. The pka = 4.76. d. Determine the pH after 50.0 mL of titrant (NaOH) have been added. This is the equivalence point. All of the acid has been converted to its conjugate base, pH is determined by the equilibrium for the conjugate base
Consider the titration of a 25.1 −mL sample of 0.125 M RbOH with 0.100 M HCl. Determine each of the following.the initial pH, the volume of added acid required to reach the equivalence point,he pH at 4.9 mL of added acid,the pH at the equivalence pointthe pH after adding 4.2 mL of acid beyond the equivalence point