Question

A sample of 447.9 mg of hydroxylamine (NH2OH) is dissolved in 55.5 mL of DI H2O....

A sample of 447.9 mg of hydroxylamine (NH2OH) is dissolved in 55.5 mL of DI H2O. This sample is to be titrated with 0.315 M HClO4 . Fill in the following table for each different point (a, b, c, etc….) during the titration. (Make sure all work is attached for credit). Also graph the titration curve with pH on the y­axis and Vbase on the x­axis.

Point |Vbase (mL)| molbase added| mol acid remain| molbase remain| [OH ­ ]| [H3O + ]| pH|

a            0.00

b           7.50

c           13.75

d           38.90

e            Veq=

f            50

Homework Answers

Answer #1
Point

VBase(mL)

Mol base added Mol acid remain Mol base remanin [OH-] [H3O+] pH
a

0.00

0.0000

0.0111 0.0000 0.0000 0.0110 1.96
b 7.50 0.0018 0.0092 0.0000 0.0000 0.0092 2.04
c 13.75 0.0034 0.0077 0.0000 0.0000 0.0077 2.12
d 38.90 0.0095 0.0015 0.0000 0.0000 0.0015 2.81
e Veq=45.08 0.0110 0.0000 0.0000 0.0000 0.0000 4.60
f 50 0.0122 0.0000 0.0012 0.0012 0.0000 11.07

Note: assume a 35mL of HClO4 because you do not know the volume of acid

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A sample of 180.0 mL of 1.65M HNO3 is titrated with 0.90M KOH. Fill in the...
A sample of 180.0 mL of 1.65M HNO3 is titrated with 0.90M KOH. Fill in the following table for each different point (a, b, c, etc….) during the titration. Also graph the titration curve with pH on the y­axis and Vbase on the x­axis (Make sure all work is attached for credit) Point |Vbase (mL)| molbase added| mol acid remain| molbase remain| [OH ­ ]| [H3O + ]| pH| a            0.00 b           75.00 c           200.25 d           275.50 e            Veq= f           ...
120 mg of morphine are dissolved in 20.0 mL water. This solution is titrated with 0.0225...
120 mg of morphine are dissolved in 20.0 mL water. This solution is titrated with 0.0225 M HCl. Calculate the pH at the equivalence point of the titration
A sample of 0.1387 g of an unknown monoprotic acid was dissolved in 25.0 mL of...
A sample of 0.1387 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.1150 MNaOH. The acid required 15.5 mL of base to reach the equivalence point. Part B After 7.25 mL of base had been added in the titration, the pH was found to be 2.45. What is the Ka for the unknown acid? (Do not ignore the change, x, in the equation for Ka.)
a. A 0.6740-g sample of impure Ca(OH)2 is dissolved in enough water to make 57.70 mL...
a. A 0.6740-g sample of impure Ca(OH)2 is dissolved in enough water to make 57.70 mL of solution. 20.00 mL of the resulting solution is then titrated with 0.2546-M HCl. What is the percent purity of the calcium hydroxide if the titration requires 17.83 mL of the acid to reach the endpoint? b. What is the iodide ion concentration in a solution if the addition of an excess of 0.100 M Pb(NO3)2 to 33.8 mL of the solution produces 565.2...
Q1: The phosphoric acid in a 100.00-mL sample of cola drink was titrated with 0.1025 N...
Q1: The phosphoric acid in a 100.00-mL sample of cola drink was titrated with 0.1025 N NaOH. If the first equivalence point occurred after 13.11 mL of base was added, and the second equivalence point occurred after 28.55 mL of base, calculate: a.     the concentrations or molarity of H3PO4           {H3PO4 + OH- à H2PO4- + H2O} b.     The concentration of H2PO4- in the cola sample. {H2PO4- + OH- à HPO4-2 + H2O} c.     How would CO2 interfere with the titration...
A 1.26 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water...
A 1.26 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water and titrated with a a 0.306 M aqueous potassium hydroxide solution. It is observed that after 12.7 milliliters of potassium hydroxide have been added, the pH is 3.193 and that an additional 19.3 mL of the potassium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid? ____ g/mol (2) What is the value of Ka...
A 0.872 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water...
A 0.872 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water and titrated with a a 0.424 M aqueous sodium hydroxide solution. It is observed that after 9.40 milliliters of sodium hydroxide have been added, the pH is 3.350 and that an additional 5.50 mL of the sodium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid? g/mol (2) What is the value of Ka for...
A 1.54 gram sample of an unknown monoprotic acid is dissolved in 30.0 mL of water...
A 1.54 gram sample of an unknown monoprotic acid is dissolved in 30.0 mL of water and titrated with a a 0.225 M aqueous potassium hydroxide solution. It is observed that after 19.3 milliliters of potassium hydroxide have been added, the pH is 7.171 and that an additional 34.5 mL of the potassium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid?  g/mol (2) What is the value of Ka for the...
A 1.22 gram sample of an unknown monoprotic acid is dissolved in 25.0 mL of water...
A 1.22 gram sample of an unknown monoprotic acid is dissolved in 25.0 mL of water and titrated with a a 0.299 M aqueous potassium hydroxide solution. It is observed that after 10.4 milliliters of potassium hydroxide have been added, the pH is 3.039 and that an additional 19.4 mL of the potassium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid? g/mol (2) What is the value of Ka for...
A 0.4016 g sample of impure sodium carbonate (soda ash) is dissolved in 50 mL of...
A 0.4016 g sample of impure sodium carbonate (soda ash) is dissolved in 50 mL of distilled water. Phenolphthalein was added and 11.40 mL of 0.09942 M HCL was required to reach the first end point. Excess volume of HCL was added according to the lab procedure (if x is the amount of acid needed to complete the first titration, you will add a total of 2x + 10mL of acid to ensure excess). After boiling, the excess acid was...