Nitric acid, HNO2 is a weak acid with Ka= 4.5 x 10-4. It dissociates according to: HNO2 (aq) H+ (aq) + NO2- (aq)
Discipline: Chemistry
Type of Paper: Question-Answer
Academic Level: Undergrad. (yrs 3-4)
Paper Format: APA
Question
(6)
HNO2 (aq) <--> H+ (aq) + NO2- (aq)
a.) What is the pH of a buffer solution which is 0.80 M in NO2- and 0.40 M in HNO2?
b.) What will be the pH of the resulting solution when 2.0 mL of 0.5 M HCl are added to 50 mL of this (from part a) buffer solution?
c.) What will be the pH of the resulting solution when 2.0 mL of 0.5 M NaOH are added to 50 mL of this (from part a) buffer solution?
An answer accompanied by a complete, step by step solution would be greatly appreciated. Thanks
(a)
Consider the dissociation reaction of nitrous acid as shown below:
Write the Henderson-Hasselbalch equation for the given reaction as shown below:
![[NO,:]
pH = pK, +log -
[HNO2]](https://media1.cheggcdn.com/media/e14/e146515e-9927-45d7-8332-9cd77476f72d/image18.png?u=74e8d097-30e4-42db-8eb9-25caf36c914a&q=d4c31806-ec90-11e3-bf69-021505f49f8a&context=gdg-qna-dgs&Policy=eyJTdGF0ZW1lbnQiOlt7IlJlc291cmNlIjoiaHR0cHM6Ly9tZWRpYTEuY2hlZ2djZG4uY29tL21lZGlhL2UxNC9lMTQ2NTE1ZS05OTI3LTQ1ZDctODMzMi05Y2Q3NzQ3NmY3MmQvaW1hZ2UxOC5wbmc~dT03NGU4ZDA5Ny0zMGU0LTQyZGItOGViOS0yNWNhZjM2YzkxNGEmcT1kNGMzMTgwNi1lYzkwLTExZTMtYmY2OS0wMjE1MDVmNDlmOGEmY29udGV4dD1nZGctcW5hLWRncyIsIkNvbmRpdGlvbiI6eyJEYXRlTGVzc1RoYW4iOnsiQVdTOkVwb2NoVGltZSI6MTY4MDYwOTE0Nn0sIkRhdGVHcmVhdGVyVGhhbiI6eyJBV1M6RXBvY2hUaW1lIjoxNjgwNjA3MDQ2fX19XX0_&Key-Pair-Id=K1M1OMQ09SPKJC&Signature=IIgzDE-bFNAZUPwXahoV9Fv5lWfKJXXut3JRHdfsTxu3939GTvv7Cr73ozNMhAAu5bWYJ-5C6e8XzENvewA~RNMuDLZTXMO0WWOEqWG81p7HYIw-QfVnQlWmRqB19YqfqBtWxl2sZed3tX64tCc-V2nKG~MTItQbQZhSk5cTgowPVvHcs2Zuc9Lj1X6Vo08P2dTuHvvv6gi0ktdaUjBv4sJrDLYg0pLvMgvgwwSnv1-lFBSM7rflPHKVmYPn3xaJ3N1Sf3CV2bPdmE22K9VAIoMWReWEFetYTJybChfv4jPxwabM~BUE77EMPYiYIzGkt6sahDzgWV~UIz-K4pcuEw__)
Consider the given data:
![K. of HNO, = 4.5*10-4
PK, of HNO2 = 3.35
The concentration of NO, NO,]=0.80M
The concentration of HNO, JHNO, 1=0.40M](https://media1.cheggcdn.com/media/cd5/cd5d2d84-eac6-4d51-89bc-aff75e5dbec7/image19.png?u=74e8d097-30e4-42db-8eb9-25caf36c914a&q=d4c31806-ec90-11e3-bf69-021505f49f8a&context=gdg-qna-dgs&Policy=eyJTdGF0ZW1lbnQiOlt7IlJlc291cmNlIjoiaHR0cHM6Ly9tZWRpYTEuY2hlZ2djZG4uY29tL21lZGlhL2NkNS9jZDVkMmQ4NC1lYWM2LTRkNTEtODliYy1hZmY3NWU1ZGJlYzcvaW1hZ2UxOS5wbmc~dT03NGU4ZDA5Ny0zMGU0LTQyZGItOGViOS0yNWNhZjM2YzkxNGEmcT1kNGMzMTgwNi1lYzkwLTExZTMtYmY2OS0wMjE1MDVmNDlmOGEmY29udGV4dD1nZGctcW5hLWRncyIsIkNvbmRpdGlvbiI6eyJEYXRlTGVzc1RoYW4iOnsiQVdTOkVwb2NoVGltZSI6MTY4MDYwOTE0Nn0sIkRhdGVHcmVhdGVyVGhhbiI6eyJBV1M6RXBvY2hUaW1lIjoxNjgwNjA3MDQ2fX19XX0_&Key-Pair-Id=K1M1OMQ09SPKJC&Signature=hYIc6YOeeahV-eaSqAvuAbPj4HL2isM-tQzNsbYqtIrw1jIgroT0TiamNd-2OHDItb5Rtr7TeL0U2J1eoLd4mKUJuYkB3M0cCxLyhQcQeQvtO8MLmxWnSYrK0lxMf8Y2Dli~fqsefWuJwCbyhUAWGY6x4OYIiDvRYbiU0ztxISrLugZSW6PGjZzxoGHON3NAf0gd3qM4g7sQD9udFMXJYKBVwfPI1Tb6fKyh5aCBXNHdovNbgvTOQ2YiWbnLaH30nzcCFtAK~in5s2I10uQu2HjELAMxnZswescsVhrntiH399e~TCzW0E8m4-SJ9E6apKBpUeVgqiO2qvjRG~~j8Q__)
Substitute the values in above equation as shown below:
![[0.80]
pH = PK, +log
[0.40]
= 3.35+0.3
= 3.65](https://media1.cheggcdn.com/media/65d/65d5375d-ca0d-42f4-87e4-d7ba61a751ba/image20.png?u=74e8d097-30e4-42db-8eb9-25caf36c914a&q=d4c31806-ec90-11e3-bf69-021505f49f8a&context=gdg-qna-dgs&Policy=eyJTdGF0ZW1lbnQiOlt7IlJlc291cmNlIjoiaHR0cHM6Ly9tZWRpYTEuY2hlZ2djZG4uY29tL21lZGlhLzY1ZC82NWQ1Mzc1ZC1jYTBkLTQyZjQtODdlNC1kN2JhNjFhNzUxYmEvaW1hZ2UyMC5wbmc~dT03NGU4ZDA5Ny0zMGU0LTQyZGItOGViOS0yNWNhZjM2YzkxNGEmcT1kNGMzMTgwNi1lYzkwLTExZTMtYmY2OS0wMjE1MDVmNDlmOGEmY29udGV4dD1nZGctcW5hLWRncyIsIkNvbmRpdGlvbiI6eyJEYXRlTGVzc1RoYW4iOnsiQVdTOkVwb2NoVGltZSI6MTY4MDYwOTE0Nn0sIkRhdGVHcmVhdGVyVGhhbiI6eyJBV1M6RXBvY2hUaW1lIjoxNjgwNjA3MDQ2fX19XX0_&Key-Pair-Id=K1M1OMQ09SPKJC&Signature=SoHzSABaQYVnPRBKL-tujuGcYwa~HaQ71peBUhh2jXXvxShJb~Uo2Gxhe~2~Jul-PS1AQlyiA~hgEAE6msdR10xb2OToCcup4-4c9oPHiQPPsxpfOkpkLjWvEbo7~z029sO0FUnG5cj2j6YUtYBQJ5u9x7jv-J8BbXXzNxPMjM2456KFVleG~ETfqXM0ZawsdOPMM5VtAjymdahlDzEhtUisG4SLtV9AAkElsULgz9T28J6viq98bYzTJGx001hGux0bqWHIGg9XLmEfFLWnZ3Q3Rh-eOAPhiFOKEHyMTlwoyd1XCzQhzbI9gdeWdhI6ySGAbYAPv~hPqmtBkwU63g__)
The pH of the buffer solution is 3.65.
is the weak acid and 
is the conjugate base of the weak acid. The concentration of is given as 0.40 M and concentration of 
is given as 0.80 M.
The dissociation constant, 
of is given as
.
Substituting all the values in the Henderson-Hasselbalch equation and the 
is obtained as, 3.65.
(b)
Consider the dissociation reaction of nitrous acid as shown below:
Consider the data shown below:
![The concentration of NO, [No; =0.80 M
The concentration of HNO,, (HNO,]=0.40M
Concentration of HCI, (HCI) =0.5M](https://media1.cheggcdn.com/media/5f0/5f0d9dc0-372c-47c9-a06f-1bf4d06b37e7/image30.png?u=74e8d097-30e4-42db-8eb9-25caf36c914a&q=d4c31806-ec90-11e3-bf69-021505f49f8a&context=gdg-qna-dgs&Policy=eyJTdGF0ZW1lbnQiOlt7IlJlc291cmNlIjoiaHR0cHM6Ly9tZWRpYTEuY2hlZ2djZG4uY29tL21lZGlhLzVmMC81ZjBkOWRjMC0zNzJjLTQ3YzktYTA2Zi0xYmY0ZDA2YjM3ZTcvaW1hZ2UzMC5wbmc~dT03NGU4ZDA5Ny0zMGU0LTQyZGItOGViOS0yNWNhZjM2YzkxNGEmcT1kNGMzMTgwNi1lYzkwLTExZTMtYmY2OS0wMjE1MDVmNDlmOGEmY29udGV4dD1nZGctcW5hLWRncyIsIkNvbmRpdGlvbiI6eyJEYXRlTGVzc1RoYW4iOnsiQVdTOkVwb2NoVGltZSI6MTY4MDYwOTE0Nn0sIkRhdGVHcmVhdGVyVGhhbiI6eyJBV1M6RXBvY2hUaW1lIjoxNjgwNjA3MDQ2fX19XX0_&Key-Pair-Id=K1M1OMQ09SPKJC&Signature=g71mV5gejD84~GTEEVcQo8JRm8jWO9qasEzFIEpak9M732G2KpVb5~SIghOZQJAN7VjMg0W4J9LNZvhsLCqt6dDNEwqq8~vN6kOn9yJpIL56x-nBU0nz38Xv9ywcSTcA9EuqsznfvrhSIynDEP-AswPa~kt1-2IQhA0rrNbdb8N9jPSoouH46SD99wrD8udhWVzpVLDlCPN3llvbP-Nz~ACJv9gLQ0OsOWsqwtG35VXdgumeqK5Tw0ZnOhvyIs7vkKFZqZmUM0qXyrmP1-xnUOZsyQ4sHJMOUqa03JQLWx5x7OcOXtHB~RW5WFnqAmlpySx6mTZaWUx9IglcVNQqCA__)
Concentration of HCl is given as 0.5 M, concentration of 
is given as 0.40 M and concentration of 
is given as 0.80 M. Volume of 
is 0.2 mL and volume of the buffer solution is 50 mL.
Moles of each species are found by multiplying molarity with volume of the solution in liters. Therefore, moles of 
, 
and HCl is found to be 0.04, 0.02 and 0.001 moles respectively.
Consider the ICE table for the change in the concentration of each species after addition of acid:
The first row is the initial moles.
The initial moles of 
is 0.001, 0.04 and 0.02 moles respectively.
The
second row is the change in moles. Assuming the net direction of the
reaction is towards the product, which gives a negative sign for
reactants and positive sign for products. Let the change be 
. Then the change in moles of 
and 
is 
mol and the change in moles of 
is 
mol.
The
third row is the equilibrium moles. It is just the initial moles added
to the change in moles. At equilibrium, the moles of is 
mol, moles of is 
mol and moles of is 
mol.
Consider the change in concentration by substituting the value of variable 
as shown below:
As the moles of 
is 
mol. Hence, 
is mol. Therefore, replacing the variable 
with ; the equilibrium concentration of is approximately zero, the equilibrium concentration of 
is 0.039 mol and the equilibrium concentration of 
is 0.021 mol.
Calculate the concentration of each species as shown below:
![[No:]
pH = pK, +log
• F10% [HNO2]
= 3.35 +loo 0.75
= 3.35+ log 0.404
= 3.35+0.27
= 3.62](https://media1.cheggcdn.com/media/b00/b00acadc-e8df-47dd-869e-a2133609756c/image77.png?u=74e8d097-30e4-42db-8eb9-25caf36c914a&q=d4c31806-ec90-11e3-bf69-021505f49f8a&context=gdg-qna-dgs&Policy=eyJTdGF0ZW1lbnQiOlt7IlJlc291cmNlIjoiaHR0cHM6Ly9tZWRpYTEuY2hlZ2djZG4uY29tL21lZGlhL2IwMC9iMDBhY2FkYy1lOGRmLTQ3ZGQtODY5ZS1hMjEzMzYwOTc1NmMvaW1hZ2U3Ny5wbmc~dT03NGU4ZDA5Ny0zMGU0LTQyZGItOGViOS0yNWNhZjM2YzkxNGEmcT1kNGMzMTgwNi1lYzkwLTExZTMtYmY2OS0wMjE1MDVmNDlmOGEmY29udGV4dD1nZGctcW5hLWRncyIsIkNvbmRpdGlvbiI6eyJEYXRlTGVzc1RoYW4iOnsiQVdTOkVwb2NoVGltZSI6MTY4MDYwOTE0Nn0sIkRhdGVHcmVhdGVyVGhhbiI6eyJBV1M6RXBvY2hUaW1lIjoxNjgwNjA3MDQ2fX19XX0_&Key-Pair-Id=K1M1OMQ09SPKJC&Signature=TqMQ~R9A2mPruiu6HU8eGn37Mz2PEVGoDNp6LIG7k-g8qDnSti6onAMEBNxffCS2O1Ij1GSk232ysF3v7lV5i5xHhTKaL-sm63vqViJP8C6A3sDFOFCnPh9dCZn-BI7mKH8rPhfRvrY6ae8pBi-FlItpz~RRKNEr3Q9AdohIzflyLdnvhH7G0E5XDLa7g7MzDqNJ5bPzYsNx5Vcf26MQnCh8NF8gm0yrc94yV1z76cQrQa77on7uUrnSmuPX4tuQYaUZJ29MOdMHFiprqM80gS--8iJmkb0CStnDez2OZ9J-2hzNZB289--Xu6rGv95mfHXSwv5jDUnzgMdTxppvoA__)
The pH of the resulting solution is 3.62.
Concentration
of the species is found by dividing the number of moles with the volume
of the solution in liters. Therefore, the concentration of 
and 
is found to be 0.78 M and 0.42 M respectively.
Substituting the value in the buffer equation, the pH is found to be 3.62.
(c)
Consider the following data:
![Concentration of NO,,[NO]=0.80M
Concentration of HNO2, (HNO3)=0.40M
Concentration of NaOH, (NaOH)=0.5M](https://media1.cheggcdn.com/media/b78/b782d22a-ff8d-4838-817c-c45758e4c7e2/image80.png?u=74e8d097-30e4-42db-8eb9-25caf36c914a&q=d4c31806-ec90-11e3-bf69-021505f49f8a&context=gdg-qna-dgs&Policy=eyJTdGF0ZW1lbnQiOlt7IlJlc291cmNlIjoiaHR0cHM6Ly9tZWRpYTEuY2hlZ2djZG4uY29tL21lZGlhL2I3OC9iNzgyZDIyYS1mZjhkLTQ4MzgtODE3Yy1jNDU3NThlNGM3ZTIvaW1hZ2U4MC5wbmc~dT03NGU4ZDA5Ny0zMGU0LTQyZGItOGViOS0yNWNhZjM2YzkxNGEmcT1kNGMzMTgwNi1lYzkwLTExZTMtYmY2OS0wMjE1MDVmNDlmOGEmY29udGV4dD1nZGctcW5hLWRncyIsIkNvbmRpdGlvbiI6eyJEYXRlTGVzc1RoYW4iOnsiQVdTOkVwb2NoVGltZSI6MTY4MDYwOTE0Nn0sIkRhdGVHcmVhdGVyVGhhbiI6eyJBV1M6RXBvY2hUaW1lIjoxNjgwNjA3MDQ2fX19XX0_&Key-Pair-Id=K1M1OMQ09SPKJC&Signature=gkKmkZI6QEfCFVa6N1mTLpd2y8-nL~aC7zZRMt3FZnrRYsF4Lz6-DqmWBg~uSlILaR7lvAvC8shVM-I7lNTHySHB2LXXyKhn9u4jNDqdZX8Cf0NF1U~Wc4HRxWCVSaEf8m73sMM9VRE-PaFYx8K0ZNZkZweZln~-LbtVVPAQOfLBq1P2o983zxMobroYkS5S8MvqMDq3lryvZKcl5w1g3K2ruzqAlfIRwq9SF-p7SVPk8eV3WCPz9aqM9BPorGgcOOjWLDx~FDgWXL47gvtof9FYJ3rjD7gQA7EWXSibJSOjP5nvxB3~5OcWBxDafOG8QtgPpy~jw6OrHYZ-xE2L4Q__)
The concentration of 
is given as 0.5M, concentration of 
is given as 0.40M and concentration of 
is given as 0.80M. Volume of 
is 0.2mL and volume of the buffer solution is 50mL.
Moles of each species are found by multiplying the molarity with volume of the solution in liters. Therefore, moles of
, and 
is found to be 0.04, 0.02 and 0.001 moles respectively.
Consider the following ICE table:
The first row is the initial moles. The initial moles of 
is 0.001, 0.04 and 0.02 moles respectively.
The
second row is the change in moles. Assuming the net direction of the
reaction is towards the reactant, which gives a negative sign for
product and positive sign for reactants. Let the change be 
. Then, the change in moles of 
and 
is 
mol and the change in moles of 
is 
mol.
The third row is the equilibrium moles. It is just the initial moles plus change in moles. At equilibrium, the moles of 
is mol, moles of 
is 
mol and moles of 
is 
mol.
Substitute the value of , i.e. moles of hydroxide and redraw the ICE table as shown below:
We know that, moles of 
is 0.001mol. Hence, is 0.001mol. Therefore, replacing the variable 
with 0.001; the equilibrium concentration of 
is approximately zero, the equilibrium concentration of 
is 0.041mol and the equilibrium concentration of 
is 0.019 mol.
Consider the following calculations:
![[NO]
pH =PK, +log
JHNO
= 3.351 10.7885
-3.35+log 0.3654
= 3.35+0.33
= 3.68](https://media1.cheggcdn.com/media/5fa/5fa6b0ac-30d7-49b4-8dfa-ea3c917cf67b/image120.png?u=74e8d097-30e4-42db-8eb9-25caf36c914a&q=d4c31806-ec90-11e3-bf69-021505f49f8a&context=gdg-qna-dgs&Policy=eyJTdGF0ZW1lbnQiOlt7IlJlc291cmNlIjoiaHR0cHM6Ly9tZWRpYTEuY2hlZ2djZG4uY29tL21lZGlhLzVmYS81ZmE2YjBhYy0zMGQ3LTQ5YjQtOGRmYS1lYTNjOTE3Y2Y2N2IvaW1hZ2UxMjAucG5nP3U9NzRlOGQwOTctMzBlNC00MmRiLThlYjktMjVjYWYzNmM5MTRhJnE9ZDRjMzE4MDYtZWM5MC0xMWUzLWJmNjktMDIxNTA1ZjQ5ZjhhJmNvbnRleHQ9Z2RnLXFuYS1kZ3MiLCJDb25kaXRpb24iOnsiRGF0ZUxlc3NUaGFuIjp7IkFXUzpFcG9jaFRpbWUiOjE2ODA2MDkxNDd9LCJEYXRlR3JlYXRlclRoYW4iOnsiQVdTOkVwb2NoVGltZSI6MTY4MDYwNzA0N319fV19&Key-Pair-Id=K1M1OMQ09SPKJC&Signature=NCOMetuTq6FqXR-sxLgHM-eJT9A5N2Osd-Z6s1PQZEI3l462Tdz0MQuyW1CnLLcA~KLGRWm~j1KTmEjS-AeNIBhSqoWLOtMSNzPOAngaU6GxJ9i0UXekPIk4mQPKNY2JbE8o6QIktHwLeEHCCOux3zTEdLipEOzTpgNRuJG3lNcIH69JgGM~A9Alb6VqxY1jPPCIjDPsGGTgoaZ5XVlt2dd8GU1jcPdRui75OrWRCxHNFiphfh9JROsYGSCO3Xf~I16XvplUxAmY4HqHdebCqN0IuK31oicQvo0FCrpIEs68RPd339VFiEPbluixGITFWJv7WOP6z2dmwdPIvu97Xw__)
The
of the buffer solution is 3.68.
The
concentration of the species is found by dividing the number of moles
with the volume of the solution in liters. Therefore, the concentration
of 
and 
is found to be 0.7885 M and 0.3654 M respectively.
Substituting the value in the buffer equation, the
is found to be 3.68.