A solution of a weak monoprotic acid requires 39.28 mL of 0.01712 M NaOH to reach...
A solution of a weak monoprotic acid requires 39.28 mL of 0.01712 M NaOH to reach the equivalence point. After addition of 17.19 mL of titrant, the pH = 3.33. What is the pKa of this acid? Express your answer with two decimal places.
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A solution of a weak monoprotic acid requires 39.28 mL of
0.01712 M NaOH to reach...
In the titration of a solution of weak monoprotic acid with a 0.1275 M solution of NaOH, the pH h...
In the titration of a solution of weak monoprotic acid with a 0.1275 M solution of NaOH, the pH half way to the equivalence point was 4.48. In the titration of a second solution of the same acid, exactly twice as much of a 0.1275 M solution of NaOH was needed to reach the equivalence point. What was the pH half way to the equivalence point in this titration?
In the titration of a solution of weak monoprotic acid with...
If the pH at the equivalence point for titration of a monoprotic weak acid with NaOH...
If the pH at the equivalence point for titration of a monoprotic weak acid with NaOH is 9.00, and 10 mL of base is required to reach the equivalence point, how would you determine the pKa of the acid? the pKa is 9.00 determine the pH after 5 mL of base is added; this is the pKa determine the pH when 20 mL of base is added; this is the pKa the pKa is -log(9)
3) A weak monoprotic acid has a pKa 6.15. 50.00 mL of an 0.1250M aqueous solution...
3) A weak monoprotic acid has a pKa 6.15. 50.00 mL of an 0.1250M aqueous solution of this weak acid is titrated with 0.1000M NaOH. a) What is the equivalence point volume and 2 equivalence point volume for this titration? Find the pH b) before the titration begins; c) after 20.00 mL of the NAOH has been added, d) after 62.50 mL of the NaOH has been added; and e) after 85.00 mL of the NAOH has been added. 4)...
1) A solution of a weak monoprotic acid of unknown concentration was titrated with 0.23 M...
1) A solution of a weak monoprotic acid of unknown concentration was titrated with 0.23 M NaOH. If a 100.-mL sample of the acid solution required exactly 10. mL of the NaOH solution to reach the equivalence point, what was the original concentration of the weak acid? 2) During the titration on problem (2B), after 5.0 mL of NaOH addition, the pH = 3.68. What is the Ka of the weak acid? please show steps i have an exam tomorrow
A student performs a titration, titrating 25.00 mL of a weak monoprotic acid, HNO2, with a...
A student performs a titration, titrating 25.00 mL of a weak monoprotic acid, HNO2, with a 1.821 M solution of NaOH. They collect data, plot a titration curve and determine that the equivalence point was reached after the addition of 24.64 mL of the base. What is concentration of NO2 at equivalence point (in M)? (give your answer to three decimal places)
In the titration of a solution of weak monoprotic acid with a 0.1800 M solution of...
In the titration of a solution of weak monoprotic acid with a 0.1800 M solution of NaOH, the pH half way to the equivalence point was 4.42 . In the titration of a second solution of the same acid, exactly twice as much of a 0.1800 M solution of NaOH was needed to reach the equivalence point. What was the pH half way to the equivalence point in this titration?
In the titration of a solution of weak monoprotic acid with a 0.1525 M solution of...
In the titration of a solution of weak monoprotic acid with a 0.1525 M solution of NaOH, the pH half way to the equivalence point was 4.40 . In the titration of a second solution of the same acid, exactly twice as much of a 0.1525 M solution of NaOH was needed to reach the equivalence point. What was the pH half way to the equivalence point in this titration?
7. + -19 points My Notes A 2.256-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 35.0...
7. + -19 points My Notes A 2.256-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 35.0 mL of this solution was titrated with 0.08152-M NaOH. The pH after the addition of 25.80 mL of base was 6.74, and the equivalence point was reached with the addition of 48.31 mL of base. a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution. mmol acid b) What...
a 25.00 ml sample of a weak acid is titrated with 0.225 M NaOH. a total...
a 25.00 ml sample of a weak acid is titrated with 0.225 M NaOH. a total of 12.20 ml of the NaOH is required to reach the equivalence point where the pH is 9.96. determine the value of pka for this weak acid
For a titration of 16.8 mL of a 0.586 M weak acid (pKa = 5.89) with...
For a titration of 16.8 mL of a 0.586 M weak acid (pKa = 5.89) with 0.51 M NaOH, what is the pH after adding 36.8 mL of titrant? (Hint: this is past the equivalence point)
In the titration of a solution of weak monoprotic acid with a 0.1275 M solution of NaOH, the pH half way to the equivalence point was 4.48. In the titration of a second solution of the same acid, exactly twice as much of a 0.1275 M solution of NaOH was needed to reach the equivalence point. What was the pH half way to the equivalence point in this titration?
In the titration of a solution of weak monoprotic acid with...
3) A weak monoprotic acid has a pKa 6.15. 50.00 mL of an 0.1250M aqueous solution of this weak acid is titrated with 0.1000M NaOH. a) What is the equivalence point volume and 2 equivalence point volume for this titration? Find the pH b) before the titration begins; c) after 20.00 mL of the NAOH has been added, d) after 62.50 mL of the NaOH has been added; and e) after 85.00 mL of the NAOH has been added. 4)...
A student performs a titration, titrating 25.00 mL of a weak monoprotic acid, HNO2, with a 1.821 M solution of NaOH. They collect data, plot a titration curve and determine that the equivalence point was reached after the addition of 24.64 mL of the base. What is concentration of NO2 at equivalence point (in M)? (give your answer to three decimal places)
7. + -19 points My Notes A 2.256-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 35.0 mL of this solution was titrated with 0.08152-M NaOH. The pH after the addition of 25.80 mL of base was 6.74, and the equivalence point was reached with the addition of 48.31 mL of base. a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution. mmol acid b) What...
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