If a researcher needs 0.50 L of 3.0 M nitrite buffer at pH 3.00, then she knows that the concentration of nitrous acid (conjugate acid) plus the concentration of nitrite anion (conjugate base) must equal 3.0 M. She must then calculate the amount of each that is necessary to give this total amount and, at the same time, yield a pH = 3.00. Using the Henderson-Hasselbach equation:
pH = pKa + log ((base)/(acid))
3.00 = 3.39 + log (NO2-/HNO2)
since NO2- = 3.0 – HNO2
3.00 = 3.39 + log ((3-HNO2)/HNO2)
solving for HNO2 you get 2.13 M, therefore NO2- = 3 – 2.13 = .87 M
I understand all this except where does the 3.39 come from? I assume it is pKa?
Sodium nitrite is a solid, so she would calculte that she needs 30.0 grams to be dissolved in 0.5 L of solution. How does she calculate the 30.0 grams??
If, for example, a solution of 10.0 M HNO2 were in the lab, she would need to use 106 ml of that solution dissolved in the same 0.5 L as the 30.0 g of NaNO2. Where do they get the 106 ml from?
Suppose she wanted to change the pH of the nitrite buffer from pH 3.00 to 2.80, an acid would have to be added to the buffer. What volume of 12 M HCL is needed to make this change? From the Hasselbah equation we know 2.80 = 3.39 + log ((.87-x)/(2.13 +x)) so x =.26, she calculates that whe would need about 11 ml of 12 M HCL. I think I got this because 2.8/.26 almost equals 11, right??
Final question: Above we found that by adding 11.0 mLof 12.0 M HCL the pH of the nitrite buffer was changed from 3.00 to 2.80. Final question: How many grams of solid sodium hydroxide would the chemist add to the buffer to change the pH from 3.00 to 3.80? Below is what I’m thinking but I could be all wrong:
3.80 = 3.00 + log ((3.80-HNO2)/HNO2
solving for HNO2 you get .52 M, therefore NO2- = 3.80 –.52 =3.28 M
How many grams?
I understand all this except where does the 3.39 come from? I assume it is pKa?
Yes, 3.39 is the pKa
Sodium nitrite is a solid, so she would calculate that she needs 30.0 grams to be dissolved in 0.5 L of solution. How does she calculate the 30.0 grams??
0.87M x 0.5L = moles NaNO2.
moles NaNO2 x molar mass NaNO2 = 30 g.
If, for example, a solution of 10.0 M HNO2 were in the lab, she would need to use 106 ml of that solution dissolved in the same 0.5 L as the 30.0 g of NaNO2. Where do they get the 106 ml from?
How many moles HNO2 do you need? M x L = 2.13 x 0.5L = 1.065 moles. Then M = moles/L and rearrange to L = moles/M = 1.065/10 = 0.1065 which rounds to 106 mL.
Suppose she wanted to change the pH of the nitrite buffer from pH 3.00 to 2.80, an acid would have to be added to the buffer. What volume of 12 M HCL is needed to make this change? From the Hasselbah equation we know 2.80 = 3.39 + log ((.87-x)/(2.13 +x)) so x =.26, she calculates that whe would need about 11 ml of 12 M HCL. I think I got this because 2.8/.26 almost equals 11, right??
0.26M x 0.5L = 0.13 moles HCl needed.
M HNO3 = moles/L or L = moles/M = 0.13/12M = 0.0108L or 10.8 mL which rounds to 11 mL.
Final question: Above we found that by adding 11.0 mLof 12.0 M HCL the pH of the nitrite buffer was changed from 3.00 to 2.80. Final question: How many grams of solid sodium hydroxide would the chemist add to the buffer to change the pH from 3.00 to 3.80? Below is what I’m thinking but I could be all wrong:
3.80 = 3.00 + log ((3.80-HNO2)/HNO2
solving for HNO2 you get .52 M, therefore NO2- = 3.80 –.52 =3.28 M
How many grams?
I wouldn't do it that way and I'll show you why. First, I went through your calcn and the math part seems to be OK; however, you aren't looking for HNO2 or nitrite. You are looking for NaOH. If we substitute your numbers we get pH = 3.39 + log(3.28/0.52) = 4.19 and you wanted 3.80 so those numbers won't get it. After a lot of looking I finally saw what is going wrong. First, you substituted 3.00 for pKa instead of 3.39. The numbers you obtained are correct for that substitution. The second error you made is substituting 3.80-x for HNO2; it should be 3.00-x for HNO2. If you substitute correctly as 3.80 = 3.39 + log(30-x)/(x) and solve for x you obtain 0.84 for HNO2 and 2.16 for NO2^-. But you CAN do it this way. At this point you must realize that the final HNO2 must be 0.84 so you add 2.13-0.84 = 1.29 = (OH). For NO2- it is 3.00-0.84 = 2.16 total which means you add 1.29 M OH.
I would do it this way:
...........HNO2 + OH^- ==>NO2^- + H2O
initial....2.13....0......0.87.......
add.................x................
change......-x.....-x......+x.......+x
equil....2.13-x .....0.....0.87+x.....
3.80 = 3.39 + log(0.87+x/2.13-x)
x = 1.29M = OH^-
1.29M x 0.5L x 40(the molar mas) = ??g.
This way the answer comes out directly in M OH to add instead of calculating HNO2 and NO2 and then subtracting from the original value to obtain the amount that must be added.
Thank you so much! I am still very shaky on this, but I understand a lot better!
To calculate the grams of solid sodium hydroxide (NaOH) needed to change the pH from 3.00 to 3.80, you need to use the Henderson-Hasselbach equation as before.
The equation you mentioned:
3.80 = 3.00 + log ((3.80-HNO2)/HNO2)
is correct. However, since you are adding NaOH to the buffer solution, you need to consider the reaction that occurs between NaOH and HNO2:
NaOH + HNO2 -> NaNO2 + H2O
This reaction consumes 1 mole of NaOH for every 1 mole of HNO2.
To calculate the number of moles of NaOH needed, you can use the equation:
(HNO2 concentration before + HNO2 concentration added) * volume = (HNO2 concentration after) * volume
(2.13 M + x) * volume = 0.52 M * volume
Simplifying the equation gives:
2.13 M + x = 0.52 M
x = 0.52 M - 2.13 M
x = -1.61 M
Since the concentration of HNO2 is negative, this means you need to add NaOH to neutralize the excess HNO2. The negative molarity value tells you the amount of HNO2 that needs to be neutralized.
To calculate the mass of NaOH needed, you can use the equation:
mass of NaOH = moles of NaOH * molar mass of NaOH
moles of NaOH = -1.61 M * 0.50 L
molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol (rounded)
Substituting these values into the equation gives:
mass of NaOH = -1.61 M * 0.50 L * 40.00 g/mol = 32.20 g (rounded)
Therefore, the chemist would need to add approximately 32.20 grams of solid sodium hydroxide to the buffer to change the pH from 3.00 to 3.80.
To answer your final question, let's calculate the amount of solid sodium hydroxide (NaOH) needed to change the pH of the nitrite buffer from 3.00 to 3.80.
First, let's understand the chemical reaction that occurs when sodium hydroxide (NaOH) reacts with nitrous acid (HNO2) in the buffer solution:
NaOH + HNO2 ⟶ NaNO2 + H2O
From the balanced equation, we can see that for every 1 mole of sodium hydroxide (NaOH) reacted, 1 mole of nitrite (NO2-) is produced.
To find the amount of NaOH needed, we need to calculate the change in concentration of nitrite.
Given that the initial concentration of nitrite (NO2-) is 3.0 M and the final concentration is 3.28 M, we can calculate the change in concentration:
Change in concentration of NO2- = Final concentration of NO2- - Initial concentration of NO2-
= 3.28 M - 3.0 M
= 0.28 M
Since the change in concentration of nitrite (NO2-) is 0.28 M, we can conclude that the same amount of sodium hydroxide (NaOH) needs to be added to the solution.
Now, to calculate the amount of NaOH in grams, we need to use the molar mass of NaOH (22.99 g/mol for sodium + 16.00 g/mol for oxygen + 1.01 g/mol for hydrogen):
Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol
= 40.01 g/mol
To find the mass of NaOH, we can use the equation:
Mass of NaOH = Moles of NaOH × Molar mass of NaOH
The moles of NaOH can be calculated using the equation:
Moles of NaOH = Change in concentration of NO2- × Volume of solution (in liters)
Given that the volume of the solution is 0.50 L, we can calculate the moles of NaOH:
Moles of NaOH = 0.28 M × 0.50 L
= 0.14 moles
Now, plugging the values into the equation, we can calculate the mass of NaOH:
Mass of NaOH = 0.14 moles × 40.01 g/mol
= 5.60 grams
Therefore, the chemist would need to add approximately 5.60 grams of solid sodium hydroxide (NaOH) to the nitrite buffer to change the pH from 3.00 to 3.80.