5. Exactly 100 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. Calculate the pH for [10 pts] the point at which 100 mL of the base has been added. 5. Exactly 100 mL...
Exactly 100 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. Calculate the pH for the point at which 100 mL of the base has been added. Show all calculations
50.0 mL of 0.090 M nitrous acid (HNO2, Ka = 7.1 x 10-4), is titrated with 0.100 M NaOH, requiring 45.0 mL of strong base to reach the equivalence point. (a) What will be the pH after 35.0 mL of NaOH have been added? (b) What will be the pH at the equivalence point? (c) What will be the pH after 60.0 mL of NaOH have been added?
1.A 25.00 ml smaple of 0.523 M nitrous acid, HNO2, solution is titrated with a 0.213 M NaOH. For HNO2, Ka = 4.0 X 10^-4 a) What is the pH before any NaOH is added? b) Write the reaction that takes place as KOH solution is added to the HNO2 solution. c) write the reaction that determines the pH at the equivialnce point. -What is the pH at the equivilance point -what is the pH at the 1/2 equivilance point...
A 15.0 mL sample of 0.21 M nitrous acid (HNO2), Kg = 4.6x10-4) is titrated with 0.15 M NaOH. Calculate the pH after the addition of the following volumes of base: a. 0.00 mL b. 10.00 mL c. 21.00 mL d. 25.00 mL
A 15.0 mL sample of 0.21 M nitrous acid (HNO2), Ka = 4.6x10-4) is titrated with 0.15 M NaOH. Calculate the pH after the addition of the following volumes of base: a. 0.00 mL b. 10.00 mL c. 21.00 mL d. 25.00 mL
2. A sample of 50 ml of nitrous acid (KA = 5.6 x 10“) is titrated with 0.070 M NaOH. The equivalence point is reached after the addition of 43.2 ml of the strong base. a) (3 marks) What is the pH of the solution at the equivalence point? b) (3 marks) What is the pH of the solution after 50 ml of NaOH has been added to the original nitrous acid solution?
24A) 40mL of 0.2 M formic acid is titrated with a strong base (NaOH= 0.5 M). Determine the pH before any base has been added. Please show steps and please explain why the answer is what it is. 24B) The 40mL 0.2 M formic acid is titrated with a 6.0mL of strong base. Here NaOH can be treated as a conjugate base and formic acid is the acid. please show steps and explain! Us (24-25, Acid/base, aqueous equilibrium) 24A) (4...
2. A sample of 65 ml of nitrous acid (KA = 5.6 x 104) is titrated with 0.075 M NaOH. The equivalence point is reached after the addition of 44.5 ml of the strong base. a) (3 marks) What is the pH of the solution at the equivalence point? b) (3 marks) What is the pH of the solution after 60 ml of NaOH has been added to the original nitrous acid solution?
A 50.0 mL sample of 0.11 M nitrous acid is titrated with 0.22 M NaOH. Calculate the pH to one decimal place when the following volumes of NaOH have been added. 0.00 mL 22.0 mL 25.0 mL 33.0 mL
An analytical chemist is titrating 241.8mL of a 1.200M solution of nitrous acid HNO2 with a 0.5200M solution of NaOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 605.5mL of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places. An analytical chemist...