2. A 350 mL buffer solution containing 0.15 M HF and 0.150 M NaF is reacted with sodium hydroxide. What mass of NaOH can this buffer neutralize before the pH rises above 4.0? If the same volume of buffer was 0.35 M in HF and 0.35 M in NaF then what mass of NaOH could be handled before the pH rises above 4.0?
Well, well, well! Looks like we have ourselves a chemistry question. Time to bring out the funny molecules!
For the first part of your question, we have a 350 mL buffer solution with 0.15 M HF and 0.150 M NaF. We want to know how much NaOH it can neutralize before the pH rises above 4.0.
So, here comes the punchline: First, we need to find the number of moles of HF in the solution. To do that, we multiply the volume (in liters) by the molarity of HF. But let's not forget that 1 L is equal to 1000 mL. Are you following still? Great!
Now, let's do some math (but don't worry, I'll keep it light):
Number of moles of HF = 0.15 moles/L * 0.350 L = 0.0525 moles
Okay, now let's move on to the number of moles of NaOH that can be neutralized by HF. The stoichiometry of the reaction is 1:1, which means that for every mole of HF, we need one mole of NaOH. Are you ready to rock and roll?
Number of moles of NaOH = 0.0525 moles
But hold your horses, my friend! We're not done yet. We need to convert moles of NaOH to grams by multiplying it with the molar mass of NaOH. So, cling on to your periodic table and let's go!
Molar mass of NaOH = 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 40.00 g/mol
Now you can finally unleash the punchline:
Mass of NaOH = 0.0525 moles * 40.00 g/mol = 2.10 grams
There you have it! This buffer solution can neutralize 2.10 grams of NaOH before the pH rises above 4.0. Mission accomplished!
Now, for the second part of your question where the buffer is 0.35 M in HF and 0.35 M in NaF. What a strong buffer that is! Let's see how much NaOH it can handle before the pH rises above 4.0.
Using the same calculations as before, we find:
Number of moles of HF = 0.35 moles/L * 0.350 L = 0.1225 moles
Number of moles of NaOH = 0.1225 moles
And, hey presto, let's convert moles to grams:
Mass of NaOH = 0.1225 moles * 40.00 g/mol = 4.90 grams
Tada! This mighty buffer can handle 4.90 grams of NaOH before the pH decides to misbehave and rise above 4.0.
I hope my chemistry comedy routine has left you smiling and with the answer you were looking for!
To answer these questions, we need to understand the concept of buffer solutions and how they work. A buffer solution consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) and helps to maintain the pH of a solution within a certain range even when an acid or base is added. In this case, the weak acid is HF, and its conjugate base is NaF.
First, let's consider the initial buffer solution with a volume of 350 mL, containing 0.15 M HF and 0.150 M NaF.
1. Determine the moles of HF and NaF in the solution:
moles of HF = (0.15 mol/L) * (0.350 L) = 0.0525 mol
moles of NaF = (0.150 mol/L) * (0.350 L) = 0.0525 mol
2. The buffer can neutralize the added NaOH through the following reaction:
HF + NaOH → NaF + H2O
3. Calculate the moles of NaOH that can be neutralized by the buffer before the pH rises above 4.0:
Since HF and NaOH react in a 1:1 ratio, the moles of NaOH that can be neutralized is also 0.0525 mol.
4. Determine the mass of NaOH using its molar mass (40.00 g/mol):
mass of NaOH = moles of NaOH * molar mass of NaOH
= 0.0525 mol * 40.00 g/mol
= 2.10 g
Therefore, a mass of 2.10 grams of NaOH can be neutralized by the buffer solution before the pH rises above 4.0.
Now, let's consider the second scenario where the volume of the buffer solution remains the same (350 mL), but the concentration of HF and NaF is increased to 0.35 M.
1. Determine the moles of HF and NaF in the solution using the new concentrations:
moles of HF = (0.35 mol/L) * (0.350 L) = 0.1225 mol
moles of NaF = (0.35 mol/L) * (0.350 L) = 0.1225 mol
2. Calculate the moles of NaOH that can be neutralized by the buffer before the pH rises above 4.0:
Again, since HF and NaOH react in a 1:1 ratio, the moles of NaOH that can be neutralized is 0.1225 mol.
3. Determine the mass of NaOH using its molar mass (40.00 g/mol):
mass of NaOH = moles of NaOH * molar mass of NaOH
= 0.1225 mol * 40.00 g/mol
= 4.90 g
Therefore, in the second scenario, a mass of 4.90 grams of NaOH can be neutralized by the buffer solution before the pH rises above 4.0.
In summary, the mass of NaOH that a buffer solution can neutralize before the pH rises above 4.0 depends on the concentration of the weak acid and its conjugate base in the solution. When the concentration of HF and NaF increases from 0.15 M to 0.35 M, the buffer can handle a larger mass of NaOH before the pH rises above 4.0 (from 2.10 g to 4.90 g).
thanks for the explanation!!!
1+1=3
:)
First calculate pH of the buffer.
pH = 3.14 + log(0.15/0.15)
pH = 3.14 based on Ka for HF of 7.2E-4. You need to use the K in your text and/or notes.
Then substitute 4.0 for pH and solve for (F^-/HF). If I didn't goof I found 7.24.
Here is the equation that goes when adding NaOH.
......HF...+....OH^- ===>F^- + H2O
begin 0.0525....0........0.0525 moles
add OH..........x..........x
final.0.525-x...0.......0.0525+x....
Now substitute into the (F^-/HF) = 7.24 and solve for x. x will be the moles of NaOH that can be added without exceeding a pH of 4.00. Then convert that many moles to grams NaOH.
The second part of the problem is done the same way.