Question

A buffer is prepared by dissolving 0.0250 mol of sodium nitrate in 250.0 ml of 0.0410 M of nitrous acid HNO2. Assume no volume change after HNO2 is dissolved. Calculate the ph of this buffer.

Answer 1

The pKa is equal to the negative logarithm to the base 10 of Ka it's basically the measure of the acid dissociation constant. do you kinda get what i'm saying?

Answer 2

here's a little more info to your question Determine molarity of acid and conjugate base

1) acid = 0.0410M
2) conj. base = 0.025/0.25 = 0.10M
pKa HNO2 = 3.40

Answer 3

it's hard for me to explain because this is backwards but here lets start off with the Henderson–Hasselbalch equation:

pH = pKa + log( [base] / [acid] )

where [base] is the concentration of the conjugate base of the weak acid, and [acid] is the concentration of the weak acid itself. So if the problem states that you have n times as much base as you have acid, you will solve for pKa:

pKa = pH - log( n )

pKa ≈ pH midway, because [acid] = [base] and log( 1 ) = 0.

Answer 4

pH = pKa + log (0.1/0.041)

Answer 5

so you don't understand kPa right? but the rest you do understand?

Answer 6

Yes

Answer 7

pKa = pH - log( n )
this is the equation to find pKa but we are missing our pH so thats what we need to figure out

Answer 8

yes

Answer 9

pH = pKa + log( [base] / [acid] )
this is how you find out pH so what is the base? and what is the acid? those are the two easiest right now im going to delete my previous answers they are a bit confusing

Answer 10

http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation

Answer 11

this is what we are dealing with

Answer 12

I did the negative log of 1.4 E ^-4.

Answer 13

I got 3.85 I dont know if Im heade in the right direction.

Answer 14

Chemistry: Principles And Reactions : A Core Text

Answer 15

is this your text?

Answer 16

your not doing it right

Answer 17

1) acid = 0.0410M
2) conj. base = 0.025/0.25 = 0.10M
pKa HNO2 = 3.40

From the Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 3.4 + log ( 0.1/0.041)
pH = 3.4 + log ( 2.439)
pH = 3.4 + 0.39
pH = 3.79

pKa = pH - log( n )

lets work backwards lets just assume the gave as the ph okay? so solve pKa

Answer 18

I looked in the book it says that pka = -log, -pka= log of ka, and 10^-pka and =10^logka=ka.

Answer 19

that works too lets do that then

Answer 20

I took the negative log of 1.4 E -4. I got 3.85.Then I took 3.85 and raised it to 10^-3.85 and got 1.41E -4 again.

Answer 21

okay just keep on doing what your doing what do you think you have to do after that? are you unsure

Answer 22

I'm sorry I'm not able to explain this well

Answer 23

Its fine

Answer 24

do you know the heylman equation?

Answer 25

http://en.wikipedia.org/wiki/Acid_dissociation_constant
http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation
http://en.wikipedia.org/wiki/PH
http://answers.yahoo.com/question/index?qid=20100717115344AA8ugwr

If you have a buffer solution that is midway to the equivalence point of a titration, pH ≈ pKa (for the titration of a weak acid with a strong base). If you are not midway, but you are still working with a buffer solution, you can use the Henderson–Hasselbalch equation:

pH = pKa + log( [base] / [acid] )

where [base] is the concentration of the conjugate base of the weak acid, and [acid] is the concentration of the weak acid itself. So if the problem states that you have n times as much base as you have acid, you will solve for pKa:

pKa = pH - log( n )

pKa ≈ pH midway, because [acid] = [base] and log( 1 ) = 0.


Determine molarity of acid and conjugate base

1) acid = 0.0410M
2) conj. base = 0.025/0.25 = 0.10M
pKa HNO2 = 3.40

From the Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 3.40 + log ( 0.1/0.041)
pH = 3.4 + log ( 2.439)
pH = 3.40 + 0.39
pH = 3.79

I have to leave I have work but I'll help you out later heres some links and pretty much everything you need to know check out the yahoo answers like i think it will be helpful i'm so so sorry i couldn't explain this to you well bye~
also really read the wikipedia ones they have all the equations and how to solve them

reply to your message is correct continue and do that!