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2. A sample of 60 ml of nitrous acid (KA= 5.6 x 104) is titrated with...
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?
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?
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?
A 15 mL sample of hydrofluoric acid, Ka=6.8x10^-4 was titrated with .475 M KOH. The equivalence point was reached after addition of 19.35 mL of base. Determine the molar concentration of the original hydrofluoric acid solution, and find the pH of the solution.
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. A solution is made by titrating 46.19 mmol (millimoles) of HA and 2.24 mmol of the strong base. Then, more strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 85.1 mL ?
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. A solution is made by titrating 9.91 mmol (millimoles) of HA and 2.66 mmol of the strong base. Then, more strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 58.2 mL ?
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH.A.A solution is made by titrating 8.00 mmol (millimoles) of HA and 2.00 mmol of the strong base. What is the resulting pH?Express the pH numerically to two decimal places.B.More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 43.0 mL?Express the pH numerically to two decimal places.
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
Titration of Weak Acid with Strong Base A certain weak acid, HA, with a Ka Value of 5.61 *10^-6, is titrated with NaOH. PART A A solution is made by mixing 8.00 mmol(millimoles) of HA and 1.00 mmol of the strong base. What is the resulting pH? express the pH numerically to two decimal places. pH = ? PART B More strong base is added until the equicalence point is reached. What is the pH of this solution at the...
3.)A certain weak acid, HA, with a Ka value of 5.61×10?6, is titrated with NaOH. Part A A solution is made by titrating 7.00 mmol (millimoles) of HA and 1.00 mmol of the strong base. What is the resulting pH? Express the pH numerically to two decimal places. Part B More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 35.0 mL ?...