Exactly 100 mL of 0.14 M nitrous acid (HNO2) are titrated with a 0.14 M NaOH solution. Calculate the pH for the following.
(a) the initial solution
(b) the point at which 80 mL of the base has been added
(c) the equivalence point
(d) the point at which 105 mL of the base has been added
2.17
To find the pH at different points during the titration, we need to consider the reaction between nitrous acid (HNO2) and sodium hydroxide (NaOH):
HNO2 + NaOH -> NaNO2 + H2O
(a) At the beginning (before any base is added), we have an acidic solution of nitrous acid. To calculate the pH, we need to find the concentration of HNO2 in the 100 mL solution.
moles of HNO2 = (concentration of HNO2) * (volume of solution)
= (0.14 M) * (0.100 L)
= 0.014 mol
Since the initial volume is 100 mL and the initial concentration of HNO2 is 0.14 M, the moles of HNO2 present in the 100 mL solution are 0.014 mol.
Assuming that HNO2 is a monoprotic acid, each mole of HNO2 reacts with one mole of water, and the volume of the solution remains constant at 100 mL.
Thus, the concentration of HNO2 in the 100 mL solution is still 0.14 M.
The pH of 0.14 M nitrous acid solution is calculated using the formula:
pH = -log10 [H+]
Since HNO2 is a weak acid, it does not dissociate completely. However, since we are assuming this is a simple calculation, we will consider it to be a strong acid.
[H+] = concentration of HNO2 = 0.14 M
Therefore, the pH of the initial solution (before any base is added) is:
pH = -log10(0.14) = 0.853
(b) At the point where 80 mL of the base has been added, we need to determine the moles of unreacted HNO2 and NaOH before calculating the pH.
moles of NaOH = (concentration of NaOH) * (volume of NaOH added)
= (0.14 M) * (0.080 L)
= 0.0112 mol
moles of HNO2 remaining = Initial moles of HNO2 - moles of NaOH
= 0.014 mol - 0.0112 mol
= 0.0028 mol
The remaining HNO2 has a volume of (100 mL - 80 mL) = 20 mL.
Thus, the concentration of HNO2 is:
(HNO2 remaining) / (volume of solution) = (0.0028 mol) / (0.020 L)
= 0.14 M
Again, since we are assuming HNO2 is a strong acid, the pH of 0.14 M nitrous acid solution is:
pH = -log10(0.14) = 0.853
Therefore, the pH at the point where 80 mL of the base has been added is still 0.853 (unchanged from the initial solution).
(c) At the equivalence point, the moles of NaOH added are equal to the moles of HNO2 present initially.
moles of NaOH = (concentration of NaOH) * (volume of NaOH added)
= (0.14 M) * (V mL)
= 0.14V mol
Since NaOH reacts with HNO2 in a 1:1 ratio, the moles of NaOH added are equal to the moles of HNO2 initially present:
0.14V = 0.014
V = 0.014 / 0.14
V = 0.10 L = 100 mL
At the equivalence point, the volume of the base added is equal to the volume of the acid initially present. Therefore, the final volume is 200 mL.
The final concentration of HNO2 is:
(0.014 mol) / (0.200 L) = 0.07 M
Using the same calculation as before, the pH of a 0.07 M nitrous acid solution is:
pH = -log10(0.07) = 1.154
Therefore, the pH at the equivalence point is 1.154.
(d) At the point where 105 mL of the base has been added, we need to determine the moles of unreacted HNO2 and NaOH before calculating the pH.
moles of NaOH = (concentration of NaOH) * (volume of NaOH added)
= (0.14 M) * (0.105 L)
= 0.0147 mol
moles of HNO2 remaining = Initial moles of HNO2 - moles of NaOH
= 0.014 mol - 0.0147 mol
= -0.0007 mol
Since we cannot have negative moles of a substance, this suggests that all the HNO2 has reacted, and we are past the equivalence point.
At this point, we have an excess of NaOH, resulting in a basic solution. The pH of a basic solution can be calculated using the formula:
pOH = -log10[OH-]
[H+] = Kw / [OH-]
= 1.0 x 10^-14 / (0.14 M)
= 7.14 x 10^-14 M
pH = -log10(7.14 x 10^-14) = 13.15
Therefore, the pH at the point where 105 mL of the base has been added is 13.15.
To calculate the pH at different points during the titration, we need to consider the dissociation of nitrous acid (HNO2) and the reaction with sodium hydroxide (NaOH) as well as the resulting concentrations of the species involved. Here's how you can solve this problem:
(a) To calculate the pH of the initial solution, we need to determine the concentration of HNO2. We are given that the initial solution has a volume of 100 mL and a concentration of 0.14 M. The concentration of HNO2 remains the same in this case because the acid has not reacted with the base yet. The pH of a weak acid like nitrous acid can be calculated using the formula:
pH = -log[H+]
Since HNO2 is a weak acid, it does not fully dissociate. Therefore, we need to calculate the concentration of H+ ions. In this case, since the concentration of HNO2 is 0.14 M, the concentration of H+ ions will also be 0.14 M. Plugging this value into the pH formula:
pH = -log(0.14) ≈ 0.853
So, the pH of the initial solution is approximately 0.853.
(b) To calculate the pH at the point where 80 mL of the base has been added, we need to determine the limiting reactant, which will then determine the concentration of the species involved.
Since both the initial concentration of HNO2 (0.14 M) and the NaOH solution (0.14 M) are the same, the limiting reactant will be the one that is present in a smaller amount.
To find the limiting reactant, we need to calculate the moles of each species.
Moles of HNO2 = concentration × volume in liters = 0.14 M × 0.08 L = 0.0112 mol
Moles of NaOH = concentration × volume in liters = 0.14 M × 0.08 L = 0.0112 mol
Since the moles are equal, we can conclude that neither HNO2 nor NaOH is a limiting reactant at this point. Thus, we have an excess of NaOH.
When NaOH reacts with HNO2, they undergo a neutralization reaction to form water (H2O) and a salt (NaNO2). This means that all the NaOH added has reacted with HNO2 to form the salt NaNO2. In this case, we have added 80 mL of NaOH, which is equal to 0.08 L.
The reaction between HNO2 and NaOH is 1:1, so the moles of HNO2 that have reacted will be the same as the moles of NaOH added:
Moles of HNO2 reacted = 0.0112 mol
Since the volume does not change (100 mL), the remaining moles of HNO2 in the solution will be the difference between the initial moles and the moles reacted:
Moles of HNO2 remaining = 0.0112 mol - 0.0112 mol = 0 mol
Since the number of moles of HNO2 remaining is 0, the concentration of HNO2 is also 0. This means that the HNO2 has been completely neutralized by the NaOH, resulting in a complete salt formation and no acidic species remaining in the solution.
Therefore, at the point where 80 mL of NaOH has been added, the solution becomes neutral (pH = 7).
(c) The equivalence point occurs when the stoichiometric amount of acid (HNO2) has reacted with the stoichiometric amount of base (NaOH). For this reaction, the stoichiometric ratio is 1:1. Since the initial concentration and volume of HNO2 are the same as NaOH, it will require the same volume of NaOH to reach the equivalence point as the volume of HNO2.
Therefore, the volume of NaOH needed to reach the equivalence point is 100 mL. At this point, all of the HNO2 has reacted with NaOH to form NaNO2. This means that the HNO2 has been completely neutralized, resulting in a complete salt formation and no acidic species remaining.
Thus, at the equivalence point, the solution becomes neutral (pH = 7).
(d) To calculate the pH at the point where 105 mL of the base has been added, we follow a similar approach as in part (b), taking into account the limiting reactant.
The moles of HNO2 reacted = volume of NaOH added × concentration of NaOH = 0.105 L × 0.14 M = 0.0147 mol
Since the stoichiometric ratio between HNO2 and NaOH is 1:1, the moles of HNO2 remaining will be the difference between the initial moles and the reacted moles.
Moles of HNO2 remaining = initial moles - moles reacted = 0.014 mol - 0.0147 mol ≈ -0.0007 mol
The negative value indicates that all the HNO2 has been consumed. In a similar manner as in part (b), this means that the HNO2 has been completely neutralized, resulting in a complete salt formation and no acidic species remaining.
Hence, at the point where 105 mL of NaOH has been added, the solution becomes neutral (pH = 7).