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.
I got pH= 11.74.
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.
Actually, when I did the problem again, I got 8.08.
Is this correct?
Yes, 8. something is right (I think I obtained 8.11 and I used 4E-4 for Ka for HNO2.
To calculate the pH at the equivalence point for the titration of HNO2 (nitrous acid) and KOH (potassium hydroxide), we need to determine the reaction that takes place and the resulting species in the solution.
The balanced chemical equation for the reaction between HNO2 and KOH is as follows:
HNO2 + KOH -> KNO2 + H2O
In this reaction, nitrous acid (HNO2) reacts with potassium hydroxide (KOH) to form potassium nitrite (KNO2) and water (H2O).
At the equivalence point, the moles of HNO2 will be equal to the moles of KOH. This means that the acid will be completely neutralized by the base. Therefore, we need to find the volume of KOH that is required to neutralize the given volume and concentration of HNO2.
To calculate the volume of KOH required, we can use the following equation:
nHNO2 = cHNO2 × V(ml)
where:
nHNO2 is the number of moles of HNO2,
cHNO2 is the concentration of HNO2 (in M),
and V is the volume of HNO2 (in mL).
Given:
cHNO2 = 0.100 M
V = 40.0 mL
nHNO2 = (0.100 M) × (40.0 mL)
= 4.00 mmol
According to the balanced equation, the reaction ratio between HNO2 and KOH is 1:1. Therefore, the number of moles of KOH required to neutralize the given amount of HNO2 is also 4.00 mmol.
Now, we need to determine the concentration of KOH in the solution at the equivalence point. To do this, we divide the moles of KOH by the volume of the solution.
Let's say that Veq is the volume of KOH required to reach the equivalence point. Since Veq is not given, we can calculate it using the following equation:
nKOH = cKOH × Veq(ml)
where:
nKOH is the number of moles of KOH,
cKOH is the concentration of KOH (in M),
and Veq is the volume of KOH at the equivalence point (in mL).
Since nKOH = 4.00 mmol and cKOH = 0.200 M, we have:
4.00 mmol = (0.200 M) × Veq
Veq = (4.00 mmol) / (0.200 M)
Veq = 20.0 mL
Therefore, at the equivalence point, the volume of KOH required is 20.0 mL.
Now, let's calculate the concentration of the species present at the equivalence point:
For KOH:
cKOH = nKOH / Veq
= (4.00 mmol) / (20.0 mL)
= 0.200 M
For KNO2 (the resulting species):
cKNO2 = cKOH
= 0.200 M
Since KNO2 is a salt formed from a strong base and a weak acid, it undergoes hydrolysis in water. The hydrolysis reaction can be written as follows:
KNO2 + H2O ⇌ KOH + HNO2
Since nitrous acid (HNO2) is a weak acid, it will partially dissociate in water, releasing H+ ions. The hydroxide ions (OH-) from the KOH will also react with the H+ ions to form water (H2O). This indicates that the solution at the equivalence point will be basic.
At the equivalence point, the concentration of OH- ions will be equal to the concentration of H+ ions formed by the hydrolysis of the nitrous acid.
Since the weak acid HNO2 is being fully neutralized, the concentration of OH- ions will be equal to the concentration of KNO2 formed. Therefore, we can calculate the concentration of OH- ions as follows:
cOH- = cKNO2
= 0.200 M
To calculate the pOH at the equivalence point, we use the equation:
pOH = -log10(cOH-)
pOH = -log10(0.200 M)
≈ -log10(2.0 x 10^-1)
≈ -(-0.70)
≈ 0.70
Since pH + pOH = 14 (at 25°C), we can calculate the pH at the equivalence point:
pH = 14 - pOH
= 14 - 0.70
= 13.30
Therefore, the pH at the equivalence point for this titration is approximately 13.30, not 11.74, as you stated.