500 ml of a buffer solution with ph=2.10 must be prepared using .4 M HNO2 and solid KNO2. The ka value of HNO2 is 4e-3.
a.) What mass of KNO2 should be added to 3 L of the HNO2 to make the buffer?
b.) What is the buffer's pH after 150 ml of .5 M HNO3 is added?
Your answer was: Approx means I've estimated. You should go through yourself and get good answers.
pKa = approx 2.4
mols HNO2 = 0.5L x 0.4M = 0.2 = acid
pH = pKa + log (base)/(acid)
2.10 = 2.40 + log (x/0.2)
Solve for x = mols KNO2 = about 0.1 mol = ? grams.
b.
millimols HNO2 = 500 mL x 0.4M = 200
mmols HNO3 added = 150 x 0.5M = 75
mmols KBNO2 = 100 from above.
............NO2^- + H^+ ==> HNO2
I.......... 100......0.......200
added........0......75........0
C...........-75.....-75.......+75
E............25......0........275
Plug that E line (after you've recalculated everything to get the good numbers to use) into HH equation and solve for pH.
I understand everything except in part a, why do you multiply the .4 M by .5 and not 3 L (since in part a it is asking for the mass you should add to 3 L of the HNO2)? Or is the 3 L just irrelevant information?
Well, it's this way. If you read the problem, as I did, the first line says you want to prepare 500 mL of a buffer. When I read that I quit reading and wrote instructions for preparing 500 mL of said buffer. You read the whole problem, which I should have done, and saw the 3L. So my question is, how much solution do you want? Do you want 500 mL? If so my instructions are good. If you want 3L, my instructions won't do it but the scheme I used will work all of them. If I had read the whole problem I would not have worked any part of it; instead I would have responded with a "which volume do you want? 500 mL or 3L?" You can blame this error on the person making the question. S/he goofed. Of couse I guess one interpretation is that s/he wants you to prepare 3 L of a buffer so you can have 500 mL but that's wasteful. Effectively you are throwing 2500 mL away to get 500 mL. Doesn't make sense to me. Taking all of this into consideration, I suspect you should use the 3L and ask the prof why the 500 mL is there. That part of the problem clearly says, "What mass of KNO2 should be added to 3 L of the HNO2 to make the buffer?
In part a, we are not multiplying the .4 M by .5. We are calculating the number of moles of HNO2 that we have in the 500 ml of solution.
In the given question, it states that we have 500 ml of a buffer solution with a pH of 2.10. The buffer solution is made using .4 M HNO2 and solid KNO2. But we need to calculate how much KNO2 is required to prepare this buffer solution.
To do this, we can use the Henderson-Hasselbalch equation, which states that pH = pKa + log ([base]/[acid]). In this equation, [base] represents the concentration of the base (KNO2) and [acid] represents the concentration of the acid (HNO2).
First, we calculate the number of moles of HNO2 we have in the 500 ml solution. We have 500 ml which is equal to 0.5 L. Using the formula Molarity (M) = moles (mol) / volume (L), we can rearrange the formula to calculate the moles: moles = Molarity x volume. So, moles HNO2 = 0.4 M x 0.5 L = 0.2 mol.
Now, we can use this value to determine the moles of KNO2 required. The Henderson-Hasselbalch equation becomes 2.10 = 2.40 + log ([KNO2]/0.2). Rearranging the equation, we get log ([KNO2]/0.2) = 2.10 - 2.40 = -0.30. Taking the antilog (10^x) of both sides gives us [KNO2]/0.2 = 0.501. Multiplying both sides by 0.2 gives us [KNO2] ≈ 0.1 mol.
To convert the moles of KNO2 to grams, we need to multiply by the molar mass of KNO2. The molar mass of KNO2 is approximately 85 g/mol. So, mass KNO2 ≈ 0.1 mol x 85 g/mol ≈ 8.5 g.
Therefore, you should add approximately 8.5 grams of KNO2 to 3 L of HNO2 to make the buffer solution.