What Volumes of HNO2 and KNO2 Create a pH 3 Buffer?

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SUMMARY

The discussion centers on calculating the volumes of 0.200 M HNO2 and 0.200 M KNO2 required to create a pH 3 buffer solution. Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), the ratio of the conjugate base (KNO2) to the acid (HNO2) is determined to be 0.398. By setting up a system of equations, specifically V_{A^-} + V_{HA} = 1L, participants conclude that the correct volumes to achieve the desired pH are 286 mL of HNO2 and 714 mL of KNO2.

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Homework Statement



. You have solutions of 0.200 M HNO2 and 0.200 M KNO2 (Ka for HNO2 = 4.00 × 10–4). A buffer of pH 3.000 is needed. What volumes of HNO2 and KNO2 are required to make 1 liter of buffered solution?
A) 500 mL of each
B) 286 mL HNO2; 714 mL KNO2
C) 413 mL HNO2; 587 mL KNO2
D) 714 mL HNO2; 286 mL KNO2
E) 587 mL HNO2; 413 mL KNO2


Homework Equations



pH=pKa+log([A]-/[HA])

The Attempt at a Solution



3=3.4+log([A-]/[HA])
-.4=log([A-]/[HA])
10^-.4
.398=[A-]/[HA]
I get confused conceptually of what to do after this. I know I need .398M of conj base for every 1M of acid
 
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Hemolymph said:
.398=[A-]/[HA]

This is one equation. The other is

V_{A^-} + V_{HA} = 1L

You need to combine volume with concentration - and you will have two equations in two unknowns. Just solve.
 

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