Answer the following two questions about the equivalence point for the titration of 27.8 mL of 0.235 M hypobromous acid (pKa = 8.64) with 0.214 M KOH.
1)Predict if the solution will be acidic, basic or neutral at the equivalence point.
2)Calculate the pH of the solution at the equivalence point.
HBrO + KOH ==> KBrO + H2O
1)What do you have at the equivalence point? That is a soln of KBrO. It is the salt of a strong base and a weak acid; therefore, the solution MUST be basic.
2) The pH at the equivalence point is determined by the hydrolysis of the salt.
............BrO^- + HOH ==> HBrO + OH^-
Set up ICE chart.
You have M x L = 27.8 x 0.235 = about .0065 moles HBrO(you need to be more exact). The equivalence point will be about 0.0065/0.214 = about 30 mL KOH.
The salt (KBrO) is then M = moles/L = 0.0065/(28+30) = about 0.11M
Substitute into Keq as follows:
Kb = (Kw/Ka) = (x)(x)/(0.11)
Solve for x, convert to pOH, then convert to pH.
To answer these questions, we need to understand the concept of the equivalence point in a titration and how to calculate the pH at that point.
1) Predicting the nature of the solution at the equivalence point:
The equivalence point is the point in a titration where the moles of acid are equal to the moles of base. For a titration between a weak acid (hypobromous acid) and a strong base (potassium hydroxide), the nature of the solution at the equivalence point depends on the strengths of the acid and the base.
Hypobromous acid (HOBr) is a weak acid, which means it does not ionize completely in water. However, potassium hydroxide (KOH) is a strong base, which completely ionizes into hydroxide ions (OH-) and potassium ions (K+) in water.
Since a weak acid and a strong base are involved, the solution will be basic at the equivalence point. This is because the excess of hydroxide ions from the strong base will make the solution basic.
2) Calculating the pH at the equivalence point:
To calculate the pH at the equivalence point, we need to determine the concentration of the salt formed when hypobromous acid reacts with potassium hydroxide.
The balanced equation for the reaction between HOBr and KOH is:
HOBr + KOH → KBr + H2O
From the balanced equation, we can see that the salt formed is potassium bromide (KBr). KBr is the conjugate base of a strong acid (HBr) and is a neutral salt.
At the equivalence point, the concentration of the potassium bromide will be determined by the molar ratio between hypobromous acid and potassium hydroxide used in the titration. In this case, the molar ratio is 1:1.
To determine the concentration of KBr, we need to calculate the moles of HOBr used:
moles of HOBr = volume of HOBr (in L) × concentration of HOBr
moles of HOBr = 27.8 mL × (1 L / 1000 mL) × 0.235 M = 0.00652 mol
Since the molar ratio is 1:1, the concentration of KBr at the equivalence point is also 0.00652 mol/L.
Since KBr is a neutral salt, it dissociates completely into its ions in water, resulting in equal concentrations of potassium ions (K+) and bromide ions (Br-).
Therefore, the concentration of hydroxide ions (OH-) at the equivalence point is 0.00652 M.
To calculate the pH at the equivalence point, we can use the fact that water undergoes autoionization:
H2O ⇌ H+ + OH-
At 25°C, the concentration of H+ and OH- in pure water is 1 × 10^-7 M (neutral pH). However, in our case, the concentration of OH- is 0.00652 M.
Using the equation:
pOH = -log10[OH-]
pOH = -log10(0.00652) ≈ 2.19
pH + pOH = 14 (at 25°C)
pH = 14 - 2.19 ≈ 11.81
Therefore, the pH of the solution at the equivalence point is approximately 11.81, indicating that the solution is basic.