#### ChemistryIsCool2013

Starting with 22.85 mL of 0.260 M nitrous acid, HNO2:
a. How many milliliters of 0.155 M potassium nitrite, KNO2, would you need to add to
produce a buffer with pH 3.60?
I have the answer just wanting to know the mechanics. When you do the hensell hasselback equ. pH=-log(pka)+log(nNO2/nHNO2-) why is the pka (normally you just plug it in) -log?

#### Unknown008

CHF Hall of Fame
Are you sure it's -log(pKa)? It should be -log(Ka) or pKa.

#### bjhopper

Buffer

Hello Chem is Cool,
Here are the mechanics.

1 Ka of Nitrous acid =4.5*10^-4
2 pKa = -logKa
3 pH = pKa +log A/HA where A = nitrite conc

#### bjhopper

Buffer

problem is double posted.See "Help with these two problems"
My answer is 64.5 ml of 0.155 M KNO2 required to add to 22.85 ml of 0.26 M HNO2solution to give 3.6 pH buffer.
Henderson-Hasselbalch equation defines the ratio of nitrite to unionized HNO2 in mols per liter in the final equilibrium mixture.Mols of nitrite to mols of HNO2 must be the same.A correction is made for the unionized HNO2.Correction for ionized nitrite from acid is too small to matter

#### Tchrgk

The pKa for HNO2 is 3.4
(0.02285L)(0.260mol/L) = 0.005941 moles HNO2
3.6 = pKa + log(NO2-/HNO2)
3.6 = 3.4 + log(x/0.005941/x)
3.6 = 3.4 + log(0.005941) - logx
logx = 3.4 - 3.6 + log(0.005941)
logx = -2.4261
x = .00375 moles NO2-
volume of NO2- = .00375mol/0.155mol/L = 0.0242L
volume of NO2- = 24.2mL

NOTE: [molNO2-/V] / [molHNO2/V] = molNO2- / molHNO2, with V = volume of the mixture. That shows that the mole ratio is equal to the molarity ratio since in the end the volume of both, NO2- and HNO2 concentrations is the same. I gave you a complete answer only because the details of this problem are tricky.

#### bjhopper

Buffer

Hello Tchrgk,
pKa = 3.35
pH-pKa = 3.6 -3.35 = 0.25
0.25 = log A/HA
10^.25 = A/HA
Mols of A /mols HA=1.8
I don't think I need to go further

#### Tchrgk

Buffer pH calculation

Good Job, BJ. Thank you for letting me know. The values of pKa's and other constants vary slightly depending on what reference source you use. The final answer will not be affected much. GK

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