One liter of a buffer composed of 1.2 M HNO2 and 0.8 M NaNO2 is mixed with 400 mL of 0.5 M NaOH. What is the new pH? Assume the pKa of HNO2 is 3.4.
I would convert to millimoles.
mmols HNO2 = mL x M = about 1200.
mmols NaNO2 = about 800.
mmols NaOH = about 200.
........HNO2 + NaOH ==> NaNO2 + H2O
I.......1200....0........800......
add...........200...............
C......-200...-200.......200.....
E.......1000...0.........1000
Plug the E line into the Hendrson-Hasselbalch equation and solve for pH
To calculate the new pH of the solution, we need to determine the concentration of the acid (HNO2) and its conjugate base (NO2-) after the reaction between HNO2 and NaOH.
Step 1: Calculate the amount of moles of HNO2 and NaNO2 in the buffer solution:
Volume of buffer solution = 1 L
Concentration of HNO2 = 1.2 M
Concentration of NaNO2 = 0.8 M
Moles of HNO2 = Concentration × Volume
= 1.2 M × 1 L
= 1.2 moles
Moles of NaNO2 = Concentration × Volume
= 0.8 M × 1 L
= 0.8 moles
Step 2: Calculate the amount of moles of NaOH added:
Volume of NaOH solution added = 400 mL = 0.4 L
Concentration of NaOH = 0.5 M
Moles of NaOH = Concentration × Volume
= 0.5 M × 0.4 L
= 0.2 moles
Step 3: Determine the limiting reagent:
HNO2 + NaOH → NaNO2 + H2O
The reaction between NaOH and HNO2 is a 1:1 ratio, so the limiting reagent is NaOH.
Step 4: Calculate the remaining moles of HNO2 and NaNO2:
Moles of HNO2 remaining = Initial moles - Moles reacted
= 1.2 moles - 0.2 moles
= 1.0 moles
Moles of NaNO2 remaining = Initial moles - Moles reacted
= 0.8 moles - 0.2 moles
= 0.6 moles
Step 5: Calculate the new concentrations of HNO2 and NO2-:
New concentration of HNO2 = Remaining moles / Volume
= 1.0 moles / 1 L
= 1.0 M
New concentration of NO2- = Remaining moles / Volume
= 0.6 moles / 1 L
= 0.6 M
Step 6: Calculate the new pH:
The Henderson-Hasselbalch equation relates the pH, pKa, and the concentrations of the acid and its conjugate base:
pH = pKa + log([A-] / [HA])
Where:
pH = the new pH
pKa = the pKa of HNO2 (3.4)
[A-] = concentration of NO2-
[HA] = concentration of HNO2
pH = 3.4 + log(0.6 / 1.0)
= 3.4 + log(0.6)
≈ 3.4 + (-0.22)
≈ 3.18
Thus, the new pH of the solution is approximately 3.18.
To calculate the new pH of the buffer solution after mixing with NaOH, we need to determine the changes to the concentrations of HNO2 and NaNO2, as well as the concentration of the hydroxide ions (OH-) from NaOH.
First, let's calculate the amount of HNO2 and NaNO2 in the original 1 liter buffer solution:
HNO2 concentration = 1.2 M
HNO2 amount in 1 liter = 1.2 mol
NaNO2 concentration = 0.8 M
NaNO2 amount in 1 liter = 0.8 mol
Next, we need to calculate the amount of NaOH added:
NaOH concentration = 0.5 M
NaOH amount added = 0.5 mol/L x 0.4 L = 0.2 mol
Since NaOH is a strong base, it will react with HNO2 (a weak acid) to form water (H2O) and NaNO2:
NaOH + HNO2 -> NaNO2 + H2O
The balanced equation shows a 1:1 stoichiometry, meaning the number of moles of HNO2 consumed will be equal to the number of moles of NaOH added (assuming complete reaction).
Therefore, the amount of HNO2 remaining in the solution will be:
HNO2 remaining = 1.2 mol - 0.2 mol = 1 mol
The amount of NaNO2 formed will be equal to the amount of NaOH added:
NaNO2 formed = 0.2 mol
Now, let's calculate the new concentration of HNO2 and NaNO2 in the final solution:
Final volume of the solution = 1 L + 0.4 L (from the added NaOH) = 1.4 L
HNO2 concentration in the final solution = HNO2 remaining / final volume
= 1 mol / 1.4 L = 0.71 M
NaNO2 concentration in the final solution = NaNO2 formed / final volume
= 0.2 mol / 1.4 L = 0.14 M
To determine the pH, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, HNO2 acts as the acid (HA), and NaNO2 acts as the conjugate base (A-) of the buffer solution.
Plugging in the known values:
pKa = 3.4
[HA] = 0.71 M (concentration of HNO2)
[A-] = 0.14 M (concentration of NaNO2)
pH = 3.4 + log(0.14/0.71) = 3.4 + log(0.197) ≈ 3.4 - 0.704 ≈ 2.696
Therefore, the new pH of the buffer solution after mixing with NaOH is approximately 2.696.