the pH if .400M sodium nitrate , NaNo2 is 8.42. what is Kb for this base?

To determine the base dissociation constant (Kb) for sodium nitrite (NaNO2), we first need to understand the relationship between pH, pOH, and the concentration of hydroxide ions in solution. Sodium nitrite is the salt of a weak acid (nitrous acid, HNO2) and a strong base (sodium hydroxide, NaOH). This means that in solution, it will hydrolyze to produce hydroxide ions, making the solution basic.

Step-by-Step Calculation

1. Calculate pOH

The pH of the solution is given as 8.42. To find the pOH, we can use the following relationship:

pH + pOH = 14

Substituting the given pH:

pOH = 14 - 8.42 = 5.58

2. Find the Concentration of Hydroxide Ions

Next, we can calculate the concentration of hydroxide ions (OH-) using the pOH:

[OH-] = 10^(-pOH)

Calculating this gives:

[OH-] = 10^(-5.58) ≈ 2.63 x 10^(-6) M

3. Set Up the Equilibrium Expression

The dissociation of sodium nitrite in water can be represented as follows:

NaNO2 + H2O ⇌ HNO2 + OH-

At equilibrium, if we let x represent the concentration of OH- produced, we can express the equilibrium concentrations:

• [HNO2] = x
• [OH-] = x
• [NaNO2] = 0.400 - x (approximately 0.400 M since x is very small)

4. Write the Kb Expression

The base dissociation constant (Kb) can be expressed as:

Kb = [HNO2][OH-] / [NaNO2]

Substituting the equilibrium concentrations into the expression:

Kb = (x)(x) / (0.400 - x)

Since x is small compared to 0.400 M, we can simplify this to:

Kb ≈ (x^2) / 0.400

5. Substitute the Value of x

Now we can substitute the value of x (which is [OH-]) into the Kb expression:

Kb ≈ (2.63 x 10^(-6))^2 / 0.400

Calculating this gives:

Kb ≈ (6.92 x 10^(-12)) / 0.400 ≈ 1.73 x 10^(-11)

Final Result

Thus, the base dissociation constant (Kb) for sodium nitrite (NaNO2) is approximately 1.73 x 10^(-11). This value indicates that sodium nitrite is a weak base, as expected from its nature as the salt of a weak acid and a strong base.