What volumes of 0.46 M HNO2 and 0.54 M NaNO2 must be mixed to prepare 1.00 L of a solution buffered at pH = 3.40?

a) HNO2
b) NaNO2
answer in L

Use the Henderson-Hasselbalch equation.

To find the volumes of 0.46 M HNO2 and 0.54 M NaNO2 needed to prepare a 1.00 L solution buffered at pH = 3.40, we need to use the Henderson-Hasselbalch equation for a buffered solution:

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

In this case, HNO2 acts as the weak acid (HA) and NaNO2 acts as its conjugate base (A-). The pKa for HNO2 is known to be 3.17. We can rearrange the Henderson-Hasselbalch equation to solve for the ratio [A-]/[HA]:

[A-]/[HA] = 10^(pH - pKa)

Substituting the given pH value (3.40) and pKa (3.17) into the equation, we get:

[A-]/[HA] = 10^(3.40 - 3.17)
[A-]/[HA] = 10^0.23
[A-]/[HA] = 1.701

The ratio [A-]/[HA] indicates that for every 1 molecule of HNO2, we need 1.701 molecules of NaNO2 in order to maintain the desired pH.

Now, we can set up an equation to solve for the volumes. Let V(HNO2) represent the volume of 0.46 M HNO2 and V(NaNO2) represent the volume of 0.54 M NaNO2. The total volume of the solution is given as 1.00 L.

V(HNO2) + V(NaNO2) = 1.00 L

Using the ratio [A-]/[HA], we have:

V(NaNO2) = V(HNO2) * [A-]/[HA]
V(NaNO2) = V(HNO2) * 1.701

Substituting this expression for V(NaNO2) into the total volume equation, we have:

V(HNO2) + V(HNO2) * 1.701 = 1.00 L
(1 + 1.701) * V(HNO2) = 1.00 L
2.701 * V(HNO2) = 1.00 L

Now, we can solve for V(HNO2):

V(HNO2) = 1.00 L / 2.701
V(HNO2) = 0.37 L (approx.)

Therefore, the volume of 0.46 M HNO2 needed to prepare the buffered solution is approximately 0.37 L.

To find the volume of NaNO2, we can substitute this value back into the equation for V(NaNO2):

V(NaNO2) = V(HNO2) * 1.701
V(NaNO2) = 0.37 L * 1.701
V(NaNO2) = 0.628 L (approx.)

Therefore, the volume of 0.54 M NaNO2 needed to prepare the buffered solution is approximately 0.628 L.