Consider the titration of 50 mL of 0.15 M NH3 with 0.1 M HCl. What is the pH at the equivalence point of the titration? (Kb= 1.8X 10-5 Ka= 5.6X10-10)

The salt present at the equivalence point is NH4Cl and there are 0.05 L x 0.15 M. That hydrolyzes at the equivalence point and that determines the pH.

NH4^+ + HOH ==> NH3^+ + H3O^+

Ka = Kw/Kb = (NH3)(H3O^+)/(NH4^+)
Plub in (NH4^+) from mols NH4Cl/liters (don't forget in liters to add the volume of NH3 initially (0.050 L) to the volume of HCl used in the titration.) and solve for H3O^+.
pH = - log (H^+). Post your work if you get stuck.

IM LOST ON THIS ONE

To find the pH at the equivalence point of the titration, we need to determine the number of moles of NH3 and HCl at the equivalence point.

Step 1: Find the number of moles of NH3:
- Volume of NH3 = 50 mL
- Molarity of NH3 = 0.15 M
- Moles of NH3 = Volume (L) x Molarity
= 50 mL x (1 L / 1000 mL) x 0.15 M
= 0.0075 moles NH3

Step 2: Find the number of moles of HCl:
- According to the balanced equation, the stoichiometric ratio between NH3 and HCl is 1:1. Therefore, at the equivalence point, the moles of HCl will be equal to the moles of NH3.
- Moles of HCl = 0.0075 moles HCl

Step 3: Calculate the concentration of NH4+ at the equivalence point:
- NH4+ is the product formed when NH3 reacts with HCl.
- Since the moles of NH4+ at the equivalence point are equal to moles of NH3 at the start, the concentration of NH4+ can be calculated using the volume of the NH3 solution.
- Volume of NH3 = Volume of NH4+ = 50 mL = 0.05 L
- Concentration of NH4+ = Moles / Volume
= 0.0075 moles / 0.05 L
= 0.15 M

Step 4: Calculate the concentration of OH- at the equivalence point:
- The reaction between NH4+ and OH- follows the equation: NH4+ + OH- ⇌ NH3 + H2O.
- Since NH4+ is a weak acid and OH- is the conjugate base of water, we can calculate OH- concentration using the Kb value.
- Kb = [NH3][OH-] / [NH4+]
= 1.8 × 10^-5
- Since [NH3] = [OH-], we can substitute [NH3] = 0.15 M (concentration of NH4+) in the equation.
- 1.8×10^-5 = (0.15)(x) / (0.15)
- x = 1.8 × 10^-5

Step 5: Calculate pOH at the equivalence point:
- pOH = -log10 [OH-]
= -log10 (1.8 × 10^-5)
= 4.74

Step 6: Calculate pH at the equivalence point:
- pOH + pH = 14 (Since pH + pOH = 14 by definition)
- pH = 14 - pOH
= 14 - 4.74
= 9.26

Therefore, the pH at the equivalence point of the titration is 9.26.

To determine the pH at the equivalence point of the titration, we need to understand the reaction that occurs between NH3 (ammonia) and HCl (hydrochloric acid).

First, let's write the balanced chemical equation for the reaction:

NH3 + HCl → NH4+ + Cl-

In this reaction, NH3 (a weak base) reacts with HCl (a strong acid) to form NH4+ (an ammonium ion) and Cl- (a chloride ion).

At the equivalence point of the titration, the moles of NH3 will react completely with the moles of HCl present. Since we know the initial concentration and volume of HCl, as well as its balanced equation stoichiometry, we can calculate the moles of HCl used.

Given:
- Initial concentration of NH3 (C1) = 0.15 M
- Volume of NH3 solution (V1) = 50 mL = 0.05 L
- Concentration of HCl (C2) = 0.1 M
- Balanced stoichiometry: 1 mole NH3 reacts with 1 mole HCl

Using the equation C × V = n (where C is concentration, V is volume, and n is the moles), we can calculate the moles of HCl used:

n(HCl) = C2 × V1
n(HCl) = 0.1 M × 0.05 L
n(HCl) = 0.005 moles

Since the moles of HCl used are equal to the moles of NH3, we can calculate the concentration of NH3 remaining at the equivalence point.
n(NH3) = n(HCl) = 0.005 moles

Now, let's calculate the concentration of NH3 (C3) at the equivalence point using the volume at the equivalence point (V3) and the remaining moles of NH3:
C3 = n(NH3) / V3

At the equivalence point, the volume of HCl solution is the same as the volume of NH3 solution (V1 = V3). Therefore, we have:
C3 = n(NH3) / V1
C3 = 0.005 moles / 0.05 L
C3 = 0.1 M

Now, we can use the given Kb value (1.8 × 10^-5) to calculate the pOH at the equivalence point. The pOH is related to the concentration of the hydroxide ion (OH-) formed as a result of the reaction between NH3 and water.

Kb = [NH4+][OH-] / [NH3]
1.8 × 10^-5 = (x)(x) / (0.1 - x)

Since NH3 is a weak base, we can assume that the concentration of OH- formed is equal to x. Substituting values, we can solve for x:

1.8 × 10^-5 = (x)(x) / (0.1 - x)
x^2 = (1.8 × 10^-5)(0.1 - x)
x^2 = 1.8 × 10^-6 - 1.8 × 10^-5x
x^2 + 1.8 × 10^-5x - 1.8 × 10^-6 = 0

Using the quadratic formula, we can solve for x (the OH- concentration) and then calculate the pOH at the equivalence point.

Once we have the pOH, we can determine the pH using the relationship:
pH = 14 - pOH

Please note that solving the quadratic equation and determining the pOH and pH at the equivalence point may require further calculations that are not shown here.