What is the pH at the equivalence point when 0.10 M HNO_3 is used to titrate a volume of solution containing 0.30 g of KOH?
HNO3 + KOH ==> KNO3 + H2O
The net ionic equation is
(H^+)(OH^-) = H2O
The pH is 7.0
To determine the pH at the equivalence point, we need to know the balanced chemical equation for the reaction between HNO3 (nitric acid) and KOH (potassium hydroxide). The balanced equation is:
HNO3 + KOH -> KNO3 + H2O
From the equation, we can see that the stoichiometric ratio between HNO3 and KOH is 1:1. This means that 1 mole of HNO3 reacts with 1 mole of KOH.
First, we need to determine the moles of KOH:
moles of KOH = mass of KOH / molar mass of KOH
The molar mass of KOH (potassium hydroxide) is calculated by summing the atomic masses of its components:
- Atomic mass of K (potassium) = 39.10 g/mol
- Atomic mass of O (oxygen) = 16.00 g/mol
- Atomic mass of H (hydrogen) = 1.01 g/mol
Molar mass of KOH = (39.10 g/mol) + (16.00 g/mol) + (1.01 g/mol) = 56.11 g/mol
Now we can calculate the moles of KOH:
moles of KOH = 0.30 g / 56.11 g/mol
Next, since the stoichiometric ratio between HNO3 and KOH is 1:1, the moles of HNO3 would also be equal to the moles of KOH.
Now, let's consider the reaction between HNO3 and KOH. At the equivalence point, all of the KOH has reacted with HNO3, and there is no excess HNO3 or KOH remaining.
This means that at the equivalence point, we have a solution containing only KNO3 (potassium nitrate) and water (H2O). The KNO3 is a neutral salt, which means it does not affect the pH of the solution.
Therefore, the pH at the equivalence point will depend on the dissociation of water. Water can undergo autoprotolysis to generate equal concentrations of H+ (hydronium ions) and OH- (hydroxide ions):
H2O(l) ⇌ H+(aq) + OH-(aq)
At 25°C, the concentrations of H+ and OH- in pure water are approximately 1.0 x 10^-7 M. However, in the presence of a strong acid (HNO3) or strong base (KOH), the concentration of H+ or OH- ions will change due to their neutralization reaction.
Since KOH is a strong base and HNO3 is a strong acid, the H+ ions from HNO3 will react with the OH- ions from KOH to form water:
H+(aq) + OH-(aq) -> H2O(l)
At the equivalence point, all the H+ ions have reacted with the OH- ions. Therefore, the concentration of H+ ions (and OH- ions) will be drastically reduced compared to their concentration in pure water.
Since the concentration of H+ ions is very low at the equivalence point, the pH will be close to 7 (neutral) but not exactly 7 due to the slight contribution of H+ ions from water autoprotolysis.
Therefore, the pH at the equivalence point when 0.10 M HNO3 is used to titrate a volume of solution containing 0.30 g of KOH will be close to 7.