How many grams of NaNO2 must be added to 1 liter of 1.00 M HNO2 solution in order to prepare a buffer solution with pH = 4.14? pH = 3.14?
My work:
ka = pH(buffer ratio)
4.5 * 10^-4 = 1.000*10^-4.14 (x M of base/1.00 M of acid)
x = 6.21
6.21 M = (6.21 moles/1 L)(1 L) = 6.21 moles
#grams NaNO2 = 6.21 moles (68.9953 g/ 1 mole) = 428 grams??
Well, well, well, looks like we've got a chemistry buff over here! But don't worry, I'm here to clown around and help you out.
First of all, congratulations on your math! Your calculations seem to be on the right track. However, we need to convert moles to grams using the molar mass of NaNO2, which is approximately 69 g/mol. So let's recalculate, shall we?
You correctly found that you need 6.21 moles of NaNO2. Now, all we have to do is multiply that by the molar mass:
6.21 moles × 69 g/mol = 428.49 grams
So, it seems like you were pretty close! But I gotta say, that's one hefty amount of NaNO2 to add to your solution. You might want to consider double-checking those calculations or finding an alternative method. Clowning around, remember? But hey, don't take my word for it - consult with a chemistry expert just to be sure!
To prepare a buffer solution with pH=4.14, you will need to add a specific amount of NaNO2 to the 1.00 M HNO2 solution. However, it seems that there is an error in your calculations.
To find the amount of NaNO2 needed, you can follow these steps:
Step 1: Calculate the concentration of the conjugate base (NO2-) required for the buffer solution using the Henderson-Hasselbalch equation:
pH = pKa + log [A-]/[HA]
4.14 = -log Ka + log [NO2-]/[HNO2]
4.14 = -log(1.0 x 10^-4.5) + log [NO2-]/1.00
Step 2: Convert the pH value to the hydrogen ion concentration [H+]:
[H+] = 10^(-pH)
[H+] = 10^(-4.14) = 7.079 x 10^-5 M
Step 3: Using the equation [HNO2] + [NO2-] = [HNO2]0 (initial concentration of acid), calculate the concentration of [NO2-]:
[HNO2] + [NO2-] = 1.00
[NO2-] = 1.00 - [HNO2]
Step 4: Substitute the calculated [NO2-] concentration into the Henderson-Hasselbalch equation from Step 1:
4.14 = -log(1.0 x 10^-4.5) + log (1.00 - [HNO2])/1.00
Step 5: Rearrange the equation to solve for [HNO2]:
log (1.00 - [HNO2])/1.00 = 4.14 + log(1.0 x 10^-4.5)
(1.00 - [HNO2])/1.00 = 10^(4.14) x 10^(-4.5)
1.00 - [HNO2] = 10^(4.14-4.5)
[HNO2] = 1.00 - (10^(4.14-4.5))
Step 6: Calculate the moles of HNO2 that is present is 1 liter of the initial solution:
[HNO2] (in moles) = concentration (in M) x volume (in L)
[HNO2] (in moles) = [HNO2] x 1.00
Step 7: Convert the moles of HNO2 into grams:
mass (in grams) = moles x molar mass
mass (in grams) = [HNO2] (in moles) x molar mass of HNO2
You can follow these steps to find the correct amount of NaNO2 required for pH = 4.14. Repeat the same calculations using pH = 3.14 to find the amount of NaNO2 required for pH = 3.14.
To calculate the amount of NaNO2 required to prepare a buffer solution with a specific pH, you need to consider the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentration ratio of the conjugate base to the weak acid:
pH = pKa + log([conjugate base]/[weak acid])
In this case, the weak acid is HNO2 and the conjugate base is NO2-. The pKa value for HNO2 can be found in a reference table or calculated. For simplicity, let's assume pKa = 3.7 (a close approximation).
pH = 4.14 when the buffer ratio is 1.00 M NO2- / 1.00 M HNO2. Plug the values into the Henderson-Hasselbalch equation:
4.14 = 3.7 + log([1.00]/[1.00])
Simplifying the equation:
0.44 = log(1/1)
Since log(1/1) = 0, we have 0.44 = 0 which is not possible. So, your calculation for pH = 4.14 seems to be incorrect.
Now, let's calculate the amount of NaNO2 required to prepare a buffer solution with pH = 3.14. Using the same equation:
3.14 = 3.7 + log([x]/[1.00])
Simplifying the equation:
-0.56 = log(x)
10^(-0.56) = x
x = 0.291 M
To convert this into grams of NaNO2, you can use the molar mass of NaNO2 (68.9953 g/mol):
grams NaNO2 = 0.291 mol x 68.9953 g/mol = 20.10 grams
Therefore, you would need approximately 20.10 grams of NaNO2 to prepare a buffer solution with pH = 3.14.