Chemistry

posted by .

If a researcher needs 0.50 L of 3.0 M nitrite buffer at pH 3.00, then she knows that the concentration of nitrous acid (conjugate acid) plus the concentration of nitrite anion (conjugate base) must equal 3.0 M. She must then calculate the amount of each that is necessary to give this total amount and, at the same time, yield a pH = 3.00. Using the Henderson-Hasselbach equation:
pH = pKa + log ((base)/(acid))
3.00 = 3.39 + log (NO2-/HNO2)
since NO2- = 3.0 – HNO2
3.00 = 3.39 + log ((3-HNO2)/HNO2)
solving for HNO2 you get 2.13 M, therefore NO2- = 3 – 2.13 = .87 M
I understand all this except where does the 3.39 come from? I assume it is pKa?



Sodium nitrite is a solid, so she would calculte that she needs 30.0 grams to be dissolved in 0.5 L of solution. How does she calculate the 30.0 grams??

If, for example, a solution of 10.0 M HNO2 were in the lab, she would need to use 106 ml of that solution dissolved in the same 0.5 L as the 30.0 g of NaNO2. Where do they get the 106 ml from?

Suppose she wanted to change the pH of the nitrite buffer from pH 3.00 to 2.80, an acid would have to be added to the buffer. What volume of 12 M HCL is needed to make this change? From the Hasselbah equation we know 2.80 = 3.39 + log ((.87-x)/(2.13 +x)) so x =.26, she calculates that whe would need about 11 ml of 12 M HCL. I think I got this because 2.8/.26 almost equals 11, right??

Final question: Above we found that by adding 11.0 mLof 12.0 M HCL the pH of the nitrite buffer was changed from 3.00 to 2.80. Final question: How many grams of solid sodium hydroxide would the chemist add to the buffer to change the pH from 3.00 to 3.80? Below is what I’m thinking but I could be all wrong:

3.80 = 3.00 + log ((3.80-HNO2)/HNO2
solving for HNO2 you get .52 M, therefore NO2- = 3.80 –.52 =3.28 M
How many grams?

  • Chemistry -

    I understand all this except where does the 3.39 come from? I assume it is pKa?

    Yes, 3.39 is the pKa

    Sodium nitrite is a solid, so she would calculate that she needs 30.0 grams to be dissolved in 0.5 L of solution. How does she calculate the 30.0 grams??

    0.87M x 0.5L = moles NaNO2.
    moles NaNO2 x molar mass NaNO2 = 30 g.


    If, for example, a solution of 10.0 M HNO2 were in the lab, she would need to use 106 ml of that solution dissolved in the same 0.5 L as the 30.0 g of NaNO2. Where do they get the 106 ml from?

    How many moles HNO2 do you need? M x L = 2.13 x 0.5L = 1.065 moles. Then M = moles/L and rearrange to L = moles/M = 1.065/10 = 0.1065 which rounds to 106 mL.

    Suppose she wanted to change the pH of the nitrite buffer from pH 3.00 to 2.80, an acid would have to be added to the buffer. What volume of 12 M HCL is needed to make this change? From the Hasselbah equation we know 2.80 = 3.39 + log ((.87-x)/(2.13 +x)) so x =.26, she calculates that whe would need about 11 ml of 12 M HCL. I think I got this because 2.8/.26 almost equals 11, right??

    0.26M x 0.5L = 0.13 moles HCl needed.
    M HNO3 = moles/L or L = moles/M = 0.13/12M = 0.0108L or 10.8 mL which rounds to 11 mL.



    Final question: Above we found that by adding 11.0 mLof 12.0 M HCL the pH of the nitrite buffer was changed from 3.00 to 2.80. Final question: How many grams of solid sodium hydroxide would the chemist add to the buffer to change the pH from 3.00 to 3.80? Below is what I’m thinking but I could be all wrong:

    3.80 = 3.00 + log ((3.80-HNO2)/HNO2
    solving for HNO2 you get .52 M, therefore NO2- = 3.80 –.52 =3.28 M
    How many grams?

    I wouldn't do it that way and I'll show you why. First, I went through your calcn and the math part seems to be OK; however, you aren't looking for HNO2 or nitrite. You are looking for NaOH. If we substitute your numbers we get pH = 3.39 + log(3.28/0.52) = 4.19 and you wanted 3.80 so those numbers won't get it. After a lot of looking I finally saw what is going wrong. First, you substituted 3.00 for pKa instead of 3.39. The numbers you obtained are correct for that substitution. The second error you made is substituting 3.80-x for HNO2; it should be 3.00-x for HNO2. If you substitute correctly as 3.80 = 3.39 + log(30-x)/(x) and solve for x you obtain 0.84 for HNO2 and 2.16 for NO2^-. But you CAN do it this way. At this point you must realize that the final HNO2 must be 0.84 so you add 2.13-0.84 = 1.29 = (OH). For NO2- it is 3.00-0.84 = 2.16 total which means you add 1.29 M OH.
    I would do it this way:
    ...........HNO2 + OH^- ==>NO2^- + H2O
    initial....2.13....0......0.87.......
    add.................x................
    change......-x.....-x......+x.......+x
    equil....2.13-x .....0.....0.87+x.....

    3.80 = 3.39 + log(0.87+x/2.13-x)
    x = 1.29M = OH^-
    1.29M x 0.5L x 40(the molar mas) = ??g.
    This way the answer comes out directly in M OH to add instead of calculating HNO2 and NO2 and then subtracting from the original value to obtain the amount that must be added.

  • Chemistry -

    Thank you so much! I am still very shaky on this, but I understand a lot better!

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry Logic

    How can I tell a conjugate base from a regular base and a conjugate acid from a regular acid?
  2. Chemistry

    The weak base-conjugate acid buffer used in this laboratory consists of a weak base ammonia, NH3, and its conjugate acid ammonium chloride, NH4Cl. If the NH3 concentration is 0.05 M and the NH4Cl concentration is 0.05 M, what is the …
  3. Chemistry

    If a liter of a buffer is prepared with a final concentration of .3M monosodium phosphate and .5M disodium phosphate, what is the pH of this buffer?
  4. chem

    Ok so for question I have to state whether the element/compound is an acid, base, conjugate acid or conjugate base. Can u please tell me if i am correct and help me out with the ones i am unsure what the answer is?
  5. chem

    Ok so i looked at it and tried to figure out the conj. base and acid. Are these right?
  6. Chemistry

    How could you prepare a buffer if only acetic acid, HCl and NaOH are available?
  7. Chemistry

    I have a question regarding the titration curve. When the ph is equal pKa that mean there are equal amount of conjugate acid to conjugate base. When a weak acid dissociates for example H3Po4 H+ H2PO4- My problem: I don’t …
  8. chemistry

    Calculate the pH of a buffer using dihydrogen phosphate and it's conjugate base. The concentration of weak acid is 0.23 M and the conjugate base concentration is 1.19 M
  9. Chemistry

    Select the statements that correctly describe buffers.?
  10. Chemistry

    In a solution of a weak acid and its conjugate base, which condition has to be true for pH to equal pKa?

More Similar Questions