A sample of 25.00 mL of 0.100 M HNO2(in a flask) is titrated with 0.150 M of NaOH solution at 25 degrees.
1) calculate the volume(Ve) of the NaOH solution needed to completely neutralize the acid in the flask.
2)calculate the pH for (a)the initial acid solution in the flask (b)the point at which 8.00 mL of the base has been added, (C) the equivalence point, and(d) the point at which 20.00 mL of the base has been added.
I can't do the entire titration for you.
1. mL acid x M acid = mL base x M base.
2. The secret to a,b,c,d, is in knowing where you are on the titration curve.
a. Pure HNO2. Before any NzOH has been added.
HNO2 ==> H^+ + NO2^-
Ka = (H^+)(NO2^-)/(HNO2)
Substitute into Ka above the following:
H^+ = x
NO2^- = x
HNO2 = 0.1 - x
Solve for x and convert to pH.
b. Calculate moles HNO2 initially.
Calculate moles NaOH at 8.00 mL
Subtract moles NaOH added from moles HNO2 initially for moles HNO2 remaining unreacted. Divide by new volume to obtain M. Determine how much of the salt was formed (in moles) and divide by the new volume to find M. Then substitute into the Ka expression and solve for H^+, then convert to pH.
c. At the equivalence point, the pH is determined by the hydrolysis of the salt.
NO2^- + HOH ==> HNO2 + OH^-
Kb for the NO2^- = (Kw/Ka) = (HNO2)(OH^-)/(NO2^-).
Prepare an ICE chart, substitute into the expression I've written and solve for OH^-, convert to pOH, then to pH.
d. The excess NaOH determines the pOH and pH. Determine how much NaOH has been added after the equivalence point, convert to moles, determine pOH, and pH.
Post your work if you get stuck.
hey did you figure this out....if so what answers did you get
To solve these questions, we will need to use stoichiometry and the concepts of acid-base reactions. Let's go step by step:
1) To calculate the volume (Ve) of NaOH solution needed to completely neutralize the acid in the flask, we can use the concept of stoichiometry. The balanced equation for the reaction between HNO2 and NaOH is as follows:
HNO2 + NaOH -> NaNO2 + H2O
From the equation, we can see that the ratio of NaOH to HNO2 is 1:1. This means that one mole of NaOH is required to neutralize one mole of HNO2.
First, let's calculate the number of moles of HNO2 in the flask. We can use the formula:
moles = concentration (M) × volume (L)
moles of HNO2 = 0.100 M × 0.025 L
moles of HNO2 = 0.0025 mol
Since the stoichiometry is 1:1, we need an equal number of moles of NaOH. Now we can calculate the volume of NaOH solution (Ve) using the formula:
Ve = moles of NaOH / concentration of NaOH
Ve = 0.0025 mol / 0.150 M
Ve = 0.0167 L = 16.7 mL
Therefore, approximately 16.7 mL of NaOH solution is needed to completely neutralize the acid in the flask.
2) Now let's calculate the pH for different points:
a) The initial acid solution in the flask:
At the beginning, we have only the 0.100 M HNO2 present. The pH of a weak acid like HNO2 can be calculated using the equation:
pH = -log[H+]
The dissociation of HNO2 can be represented as:
HNO2 ⇌ H+ + NO2-
Since HNO2 is a weak acid, it does not completely dissociate. To calculate the pH, we need to consider the initial concentration of HNO2 and the dissociation constant (Ka) of HNO2, which is 4.5 × 10^-4.
Using the formula for the pH of a weak acid,
[H+] = √(Ka × [HNO2])
[H+] = √(4.5 × 10^-4 × 0.100)
[H+] = √(4.5 × 10^-5)
[H+] ≈ 0.00671 M
Now, calculate the pH using the value of [H+]:
pH = -log(0.00671)
pH ≈ 2.17
Therefore, the pH of the initial acid solution in the flask is approximately 2.17.
b) At the point where 8.00 mL of the base has been added:
Since both the acid and base react in a 1:1 ratio, half of the acid has been neutralized when 8.00 mL of the base is added.
The moles of acid neutralized = (moles of HNO2) × (volume of base added / total volume of acid)
Since we have initially 0.025 L (25.00 mL) of the acid solution,
(moles of HNO2) = 0.0025 mol
(volume of base added / total volume of acid) = (8.00 mL) / (25.00 mL) = 0.32
(moles of acid neutralized) = 0.0025 mol × 0.32 ≈ 0.0008 mol
The remaining moles of acid = (initial moles of acid) - (moles of acid neutralized)
Remaining moles of acid = 0.0025 mol - 0.0008 mol ≈ 0.0017 mol
Now, using the remaining moles of acid and the total volume of the solution, we can calculate the concentration of the remaining acid:
(concentration of the remaining acid) = (remaining moles of acid) / (total volume of solution)
(concentration of the remaining acid) = 0.0017 mol / 0.033 L ≈ 0.05152 M
Since we know that pH = -log[H+], we can calculate the pH of the remaining acid using the concentration of the remaining acid:
pH = -log(0.05152)
pH ≈ 1.29
Therefore, at the point where 8.00 mL of the base has been added, the pH is approximately 1.29.
c) At the equivalence point:
At the equivalence point, the moles of acid and base are equal since they react in a 1:1 stoichiometric ratio. Therefore, all the HNO2 will be neutralized.
The moles of NaOH added at the equivalence point = (total volume of NaOH added) × (concentration of NaOH)
moles of NaOH added = 0.0167 L × 0.150 M = 0.0025 mol
Since the stoichiometry is 1:1, the moles of HNO2 neutralized at the equivalence point is also 0.0025 mol. This means that all of the acid has reacted, resulting in a neutral solution.
In a neutral solution, the concentration of H+ ions is equal to the concentration of OH- ions. Therefore, the pH at the equivalence point is 7 (neutral pH).
d) At the point where 20.00 mL of the base has been added:
By using the same method as in part (b), we can find the remaining moles of acid and the concentration of the remaining acid after adding 20.00 mL of the base.
The remaining moles of acid = (initial moles of acid) - (moles of acid neutralized)
Remaining moles of acid = 0.0025 mol - (0.0025 mol × (20.00 mL / 25.00 mL))
Remaining moles of acid ≈ 0.0012 mol
The remaining concentration of acid = (remaining moles of acid) / (total volume of solution)
(concentration of the remaining acid) = 0.0012 mol / 0.045 L ≈ 0.02667 M
pH = -log(0.02667)
pH ≈ 1.575
Therefore, at the point where 20.00 mL of the base has been added, the pH is approximately 1.575.