Suppose you are assigned to prepare a buffer of 0.10M HNO2 (Ka=1.7 x 10^-4) and 0.10M NaNO2 by mixing equal volumes of both solutions. What is the pH of the resultant solution?
How would I first start this question? I have no clue what the steps are to be frank.
if the volumes of the weak acid and its conjugate base are equal than the pH=pka of the solution. You know this by looking at the Henderson–Hasselbalch equation.
pH=pka+log[A^-]/[HA]
Where
A^-=NaNO2
and
HA= HNO2
Since the concentrations are equal, the above equation becomes the following:
pH=pka+log[1]
The log of 1 is 0, so the equation becomes
pH=pka
Where
pka=-log[Ka]
You can do the rest.
To determine the pH of the resultant solution, we need to understand the properties and behavior of the acid (HNO2) and its conjugate base (NO2-).
First, let's start by writing the balanced chemical equation for the dissociation of HNO2 in water:
HNO2 ⇌ H+ + NO2-
We can see that HNO2 donates a proton (H+) to water, forming the hydronium ion (H3O+), and the conjugate base (NO2-). Since the initial concentrations of HNO2 and NaNO2 are both 0.1 M, the concentrations of H+ and NO2- will also be 0.1 M.
Next, we need to calculate the concentration of HNO2 that has dissociated into H+. The dissociation constant (Ka) of HNO2 is given as 1.7 x 10^-4. The Ka value is calculated using the equation:
Ka = [H+][NO2-] / [HNO2]
We can rearrange this equation to solve for [H+]:
[H+] = (Ka * [HNO2]) / [NO2-]
Substituting the known values, we have:
[H+] = (1.7 x 10^-4 * 0.10) / 0.10
[H+] = 1.7 x 10^-4 M
Now, we have determined the concentration of H+ ions in the solution. To find the pH, we can use the equation:
pH = -log[H+]
Plugging in the concentration of H+ ions, we get:
pH = -log(1.7 x 10^-4)
Using a calculator, solve for the pH. The result will be approximately 3.77.
Therefore, the pH of the resultant solution is 3.77.