A 0.150 M solution of nitrous acid (HNO2) is made. Ka = 4.5 x 10-4

1. Show the equilibrium which occurs when this acid is dissolved in water.
2. What is the pH of the solution? Show all work clearly.
3. 100.0 mL of the solution is titrated against 0.150 M NaOH. What is the pH at the equivalence point?
4. What is true about the pH and pKa, and the [acid] and [conjugate base] at the half-equivalence point?

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1. The equilibrium which occurs when nitrous acid (HNO2) is dissolved in water can be represented by the following equation:

HNO2 + H2O ⇌ H3O+ + NO2-

In this equation, HNO2 (nitrous acid) donates a proton (H+) to water, resulting in the formation of H3O+ (hydronium ion) and NO2- (nitrite ion), which is the conjugate base of nitrous acid.

2. To find the pH of the solution, we need to determine the concentration of hydronium ions (H3O+). Since HNO2 is a weak acid, we can assume that it dissociates partially. We can use the given Ka (acid dissociation constant) value to calculate the concentration of H3O+.

Ka = [H3O+][NO2-] / [HNO2]

Since we have a 0.150 M solution of HNO2, we can assume x moles of HNO2 dissociate into x moles of H3O+ and x moles of NO2-. This means that [H3O+] and [NO2-] will both be equal to x.

Substituting these values into the Ka expression:

4.5 x 10^-4 = x^2 / (0.150 - x)

Since x is much smaller than 0.150, we can assume that 0.150 - x ≈ 0.150. Simplifying the equation:

4.5 x 10^-4 ≈ x^2 / 0.150

Rearranging:

x^2 = 4.5 x 10^-4 * 0.150

x^2 = 6.75 x 10^-5

Taking the square root:

x ≈ 8.2 x 10^-3 M

Since [H3O+] = x, the concentration of hydronium ions in the solution is approximately 8.2 x 10^-3 M.

Using the equation pH = -log[H3O+], we can calculate the pH:

pH = -log(8.2 x 10^-3)
pH ≈ 2.09

Therefore, the pH of the solution is approximately 2.09.

3. At the equivalence point, the number of moles of NaOH added will be equal to the number of moles of HNO2 initially present in the 100.0 mL solution. The reaction between HNO2 and NaOH can be represented as follows:

HNO2 + NaOH → NaNO2 + H2O

Since NaOH is a strong base and HNO2 is a weak acid, the reaction will go to completion, resulting in the formation of NaNO2 (sodium nitrite) and water (H2O). Therefore, at the equivalence point, we will have a solution of NaNO2.

The pH of the solution at the equivalence point will depend on the properties of the resulting salt, NaNO2. NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2). Therefore, it will hydrolyze to a small extent, producing hydroxide ions (OH-) and slightly increasing the pH of the solution.

Since the concentration of NaNO2 is 0.150 M and NaNO2 will produce an equal concentration of OH-, we can use the equation:

Kw = [H3O+][OH-] = 1.0 x 10^-14

Since [H3O+]+[OH-]=1 x 10^-14, we can assume that [H3O+] ≈ [OH-].

Substituting [OH-] with x:

x^2 = 1.0 x 10^-14

Taking the square root:

x ≈ 1.0 x 10^-7 M

The concentration of hydroxide ions at the equivalence point is approximately 1.0 x 10^-7 M. Therefore, the pOH of the solution is:

pOH = -log(1.0 x 10^-7)
pOH ≈ 7.0

Using the equation pH + pOH = 14, we can calculate the pH at the equivalence point:

pH + 7.0 = 14
pH ≈ 14 - 7.0
pH ≈ 7.0

Therefore, the pH of the solution at the equivalence point is approximately 7.0.

4. At the half-equivalence point, half of the initial amount of HNO2 has reacted with NaOH. This means that the concentrations of HNO2 and its conjugate base (NO2-) will be equal.

The half-equivalence point occurs when [HNO2] = [NO2-]. Furthermore, at this point, the concentration of hydronium ions ([H3O+]) will be equal to the concentration of hydroxide ions ([OH-]), resulting in a neutral solution.

The pH at the half-equivalence point will be 7, indicating neutrality. Moreover, the pKa (acid dissociation constant) is the negative logarithm of Ka, which represents the strength of an acid. At the half-equivalence point, the [acid] and [conjugate base] will be equal, making the solution a buffer. In a buffer solution, the pH is equal to the pKa.

Therefore, at the half-equivalence point, the pH will be 7, and the pKa and the concentrations of the acid and the conjugate base will be equal.