A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the equivalence point for the titration of HNO2 and KOH.

When I did this problem again, I got
pH= 8.08.

Is this close to your answer?

Dr.Bob:
I estimated the pH and didn't come close to 11 or 12 but I may not have used the same Ka value you did. Most tables in text books don't agree. Post your work and I'll look at it.

My response below at the original post.

To calculate the pH at the equivalence point for the titration of HNO2 and KOH, we need to consider the reaction that occurs between them.

The reaction between HNO2 (nitrous acid) and KOH (potassium hydroxide) is as follows:

HNO2 + KOH → KNO2 + H2O

In this reaction, HNO2 acts as a weak acid, and KOH acts as a strong base. At the equivalence point, all the HNO2 will react with KOH, resulting in the formation of KNO2 and water.

Since KNO2 is a salt, it will fully dissociate in water to give K+ and NO2- ions. The NO2- ions can react with water to form HNO2 and OH- ions through the following equilibrium reaction:

NO2- + H2O ⇌ HNO2 + OH-

The equilibrium constant for this reaction is given by the acid dissociation constant (Ka) for nitrous acid, which is equal to 4.5 x 10^-4.

At the equivalence point, the concentration of HNO2 is zero, and the concentration of OH- ions is determined by the reaction between NO2- and water. We can use the equilibrium expression for this reaction to calculate the concentration of OH- ions at the equivalence point:

[OH-] = sqrt(Ka * [NO2-])

Given that we have a 0.200 M solution of KOH, the concentration of NO2- ions will be equal to the concentration of KOH, which is 0.200 M.

[OH-] = sqrt((4.5 x 10^-4) * (0.200))

[OH-] ≈ 0.02 M

Since the concentration of OH- ions is known, we can use the pOH scale to determine the pOH at the equivalence point:

pOH = -log10([OH-])

pOH ≈ -log10(0.02)

pOH ≈ 1.70

Finally, to calculate the pH at the equivalence point, we can use the relationship between pH and pOH:

pH = 14 - pOH

pH ≈ 14 - 1.70

pH ≈ 12.30

Therefore, based on the calculations, the pH at the equivalence point for the titration of HNO2 and KOH is approximately 12.30.

It seems that your answer of pH = 8.08 is not close to the calculated value of pH = 12.30. There may have been an error in your calculations or you may have used different values for the equilibrium constant. I would recommend double-checking your calculations and using the correct values to obtain the accurate pH at the equivalence point.

To calculate the pH at the equivalence point of the titration between HNO2 and KOH, we need to consider the reaction that occurs during the titration:

HNO2 + KOH -> KNO2 + H2O

At the equivalence point, the moles of acid (HNO2) are equal to the moles of base (KOH). This means that all of the HNO2 has reacted with KOH to form KNO2. The solution is now a mixture of KNO2 and its conjugate acid, HNO2.

To find the pH at the equivalence point, we need to determine the concentration of the conjugate acid, HNO2. We can do this using the initial information provided:

Volume of HNO2 solution = 40.0 mL = 0.040 L
Concentration of HNO2 solution = 0.100 M

Using the known volume and concentration, we can calculate the moles of HNO2:

moles of HNO2 = Volume * Concentration
= 0.040 L * 0.100 M
= 0.004 moles

Since the reaction is 1:1 between HNO2 and KOH, we know that 0.004 moles of KOH were also used in the reaction.

Now, we need to determine the total volume of the solution at the equivalence point. This can be found by summing the volumes of HNO2 and KOH solutions:

Total volume at equivalence point = Volume of HNO2 + Volume of KOH
= 0.040 L + 0.040 L
= 0.080 L

Since the concentrations of HNO2 and KOH are given, and we know the moles of both, we can calculate the final concentrations of KNO2 and HNO2 in the solution at the equivalence point:

Concentration of KNO2 = moles of KNO2 / Total volume at equivalence point
= 0.004 moles / 0.080 L
= 0.050 M

Concentration of HNO2 = moles of HNO2 / Total volume at equivalence point
= 0.004 moles / 0.080 L
= 0.050 M

At the equivalence point, KNO2 and HNO2 are a conjugate acid-base pair. The pH of a solution containing a conjugate acid-base pair is determined by the pKa (or pKb) of the acid (or base). Since HNO2 is a weak acid, we can use the Henderson-Hasselbalch equation to calculate the pH:

pH = pKa + log([conjugate base]/[weak acid])

The pKa of HNO2 is not given, so we need to look it up. Let's assume it is 3.17 based on a reliable source.

Using the values calculated above, we can now calculate the pH at the equivalence point:

pH = pKa + log([0.050 M]/[0.050 M])
= 3.17 + log(1)
= 3.17

Therefore, my calculated pH at the equivalence point is 3.17, which is significantly different from the pH of 8.08 as obtained by you. There might be an error in either the calculations or some of the values used. Please double-check your work and make sure to use accurate pKa and concentration data.