Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 0.080 M NaOH.
a)hydrobromic acid (HBr)
b)chlorous acid (HClO2)
c)benzoic acid (C6H5COOH)
To calculate the pH at the equivalence point in titrating a strong acid with a strong base, we need to identify the strong acid and strong base involved in each case.
a) Hydrobromic acid (HBr) is a strong acid, and NaOH is a strong base.
b) Chlorous acid (HClO2) is a weak acid, and NaOH is a strong base.
c) Benzoic acid (C6H5COOH) is a weak acid, and NaOH is a strong base.
Since HBr is a strong acid, it will be completely ionized in water:
HBr → H+ + Br-
At the equivalence point, the moles of HBr will react stoichiometrically with the moles of NaOH. As a result, the concentration of H+ will be equal to the concentration of OH-. Therefore, the pH at the equivalence point will be equal to 7, which is neutral.
For chlorous acid (HClO2) and benzoic acid (C6H5COOH), we need to consider their acid dissociation constants (Ka) to calculate the pH at the equivalence point.
The balanced equation for the reaction between chlorous acid and sodium hydroxide is:
HClO2 + NaOH → NaClO2 + H2O
The balanced equation for the reaction between benzoic acid and sodium hydroxide is:
C6H5COOH + NaOH → C6H5COONa + H2O
To calculate the pH at the equivalence point, we need to determine the moles of each acid and base at that point.
Let's calculate the moles for each case:
a) HBr:
Moles of HBr = 0.100 M × Volume (Liters)
b) HClO2:
Moles of HClO2 = 0.100 M × Volume (Liters)
c) C6H5COOH:
Moles of C6H5COOH = 0.100 M × Volume (Liters)
Once we know the moles of the acid, we can determine the concentration of H+ or OH- at the equivalence point using stoichiometry:
For HClO2:
[H+] = Moles of HClO2 / Volume (Liters)
For C6H5COOH:
[H+] = Moles of C6H5COOH / Volume (Liters)
However, it is important to note that at the equivalence point, the concentration of OH- will be equal to the concentration of H+. Therefore, [OH-] can also be calculated using the same equations.
Once we have the concentration of H+ or OH-, we can calculate the pH using the formula:
pH = -log [H+]
To calculate the pH at the equivalence point in titrating these acids with NaOH, we need to identify the respective acid-base reactions and determine the concentration of the resulting salt solution. From this, we can find the pH using the appropriate equations.
Let's go through each case step by step:
a) Hydrobromic acid (HBr)
The acid-base reaction between HBr and NaOH is as follows:
HBr + NaOH → NaBr + H2O
The equivalence point is reached when the moles of HBr equal the moles of NaOH. Since the concentration of HBr is given as 0.100 M and the concentration of NaOH is 0.080 M, we can use the following equation to determine the concentration of the resulting NaBr salt:
moles of HBr = moles of NaOH
0.100 M × volume of HBr solution = 0.080 M × volume of NaOH solution
Since the concentrations are the same at the equivalence point, the salt concentration will be twice the initial concentration of either HBr or NaOH. Therefore, the concentration of NaBr will be:
[NaBr] = 2 × 0.100 M = 0.200 M
To find the pH, we need to consider the hydrolysis of NaBr. NaBr is a salt of a strong base (NaOH) and a strong acid (HBr), so its solution will be neutral (pH = 7) at the equivalence point.
b) Chlorous acid (HClO2)
The acid-base reaction between HClO2 and NaOH is as follows:
HClO2 + NaOH → NaClO2 + H2O
Similar to the previous case, the equivalence point is reached when the moles of HClO2 equal the moles of NaOH. Using the given concentrations, we can write the equation:
0.100 M × volume of HClO2 solution = 0.080 M × volume of NaOH solution
Again, the concentration of NaClO2 at the equivalence point will be twice the initial concentration of either HClO2 or NaOH. Thus:
[NaClO2] = 2 × 0.100 M = 0.200 M
To find the pH, we need to consider the hydrolysis of NaClO2. NaClO2 is a salt of a weak acid (HClO2) and a strong base (NaOH). The hydrolysis of ClO2- will result in the solution being slightly basic. To calculate the pH, we can use the equation:
pH = 14 - pOH
To find pOH, we need to determine the OH- concentration resulting from the hydrolysis of NaClO2. Using the equation:
Kw = [H3O+][OH-]
At the equivalence point, [H3O+] = [OH-] due to neutralization. Thus, we can write:
Kw = [OH-]^2
From this, we can find pOH:
pOH = -log([OH-])
Knowing that K_w = 1.0 × 10^-14, we can solve for [OH-] and calculate the pH using the equation above.
c) Benzoic acid (C6H5COOH)
The acid-base reaction between benzoic acid and NaOH is as follows:
C6H5COOH + NaOH → NaC6H5COO + H2O
Similarly, the equivalence point is reached when the moles of benzoic acid equals the moles of NaOH. Using the given concentrations, we can write the equation:
0.100 M × volume of benzoic acid solution = 0.080 M × volume of NaOH solution
The concentration of the resulting salt NaC6H5COO will be twice the initial concentration of either benzoic acid or NaOH:
[NaC6H5COO] = 2 × 0.100 M = 0.200 M
To calculate the pH, we need to consider the hydrolysis of NaC6H5COO. It is a salt of a weak acid (benzoic acid) and a strong base (NaOH). The hydrolysis of C6H5COO- will result in the solution being slightly basic. To calculate the pH, we can use the same procedure as explained in the case of chlorous acid (HClO2) above.
By following these steps, you can determine the pH at the equivalence point for each of the given acids.
To do these you must be able to recognize which anions/cations act as bases/acids in aqueous solution.
b part. NaClO is the salt. The ClO^- hydrolyzes in water as follows:
.........ClO^- + HOH --> HClO + OH^-
initial.0.0444.............0.....0
change...-x................x.....x
equil..0.044-x..............x....x
Kb for ClO^- = (Kw/Ka for HClO) = (HClO)(OH^-)/(ClO^-)
Substitute from ICE chart above and solve for x = (OH^-) and convert to pH.
The c part is just like the b part. For the a part, neither anion nor cation is hydrolyzed; therefore, the pH = 7.
Note: There is nothing in the problem that tells you how to handle the concn of the salt at the equivalence point but it is 0.0444 M in this instance. I don't know what level you are but it seems a little advanced for beginners.
You do it this way. Take ANY volume acid, calculate the volume of the 0.08M base it will take to neutralize it, then moles salt (M x L) divided by TOTAL volume. It makes no difference what volume you take it will always end up 0.0444M for the 0.1M/0.08M case.
HClO2 and NaOH gives you NaClO2