Answer: Let's tackle each question step by step:
Calculate the pH at the Equivalent Point of the Titration:
At the equivalent point of the titration, the moles of acid (HCl) will react completely with the moles of base (NH3) present in the solution.
Given:
Volume of NH3 solution (V1) = 40.0 mL = 0.0400 L
Molarity of NH3 solution (M1) = 0.240 M
Volume of HCl solution (V2) = unknown (since it's a titration, we need to find it)
Molarity of HCl solution (M2) = 0.400 M
Kb of NH3 = 1.8 × 10⁻⁵
At the equivalent point, the moles of NH3 (base) = moles of HCl (acid):
Moles of NH3 = M1 * V1
Moles of HCl = M2 * V2
Since the moles are equal at the equivalent point, we can set up the equation:
M1 * V1 = M2 * V2
Now, solve for V2:
V2 = (M1 * V1) / M2
V2 = (0.240 M * 0.0400 L) / 0.400 M
V2 = 0.0240 L = 24.0 mL
Now, we have the volume of HCl required to reach the equivalent point. We can use this information to calculate the moles of HCl that reacted:
Moles of HCl = M2 * V2
Moles of HCl = 0.400 M * 0.0240 L
Moles of HCl = 0.00960 moles
Now, the moles of NH3 in the original solution (before titration) can be calculated:
Moles of NH3 = M1 * V1
Moles of NH3 = 0.240 M * 0.0400 L
Moles of NH3 = 0.00960 moles
Since the moles of NH3 and HCl are equal at the equivalent point, all the NH3 will be neutralized to form NH4+ ions. Now, calculate the concentration of NH4+ ions:
Concentration of NH4+ = Moles of NH4+ / Volume of the solution at the equivalent point
Concentration of NH4+ = 0.00960 moles / 0.0400 L = 0.240 M
Now, calculate the pOH at the equivalent point:
pOH = -log(0.240) ≈ 0.62
Finally, calculate the pH at the equivalent point:
pH = 14 - pOH = 14 - 0.62 ≈ 13.38
Solubility of Calcium Phosphate (Ca3(PO4)2):
The solubility product constant (Ksp) for calcium phosphate (Ca3(PO4)2) is given as Kw(Ca3(PO4)2) = 2.07 × 10⁻⁵³.
The solubility product expression for calcium phosphate is:
Ksp = [Ca²⁺]³ [PO₄³⁻]²
Since calcium phosphate has a 1:3 molar ratio of calcium ions (Ca²⁺) to phosphate ions (PO₄³⁻), we can represent the concentration of phosphate ions as 3x and the concentration of calcium ions as x. Now, rewrite the Ksp expression:
Ksp = (x)³ * (3x)²
Now, substitute the given Ksp value and solve for x:
2.07 × 10⁻⁵³ = (x)³ * (3x)²
Solving for x, we get x ≈ 3.39 × 10⁻⁸ M
Since x represents the concentration of calcium ions (Ca²⁺), the solubility of calcium phosphate is approximately 3.39 × 10⁻⁸ M.
pH of the Solution after Mixing NaOH and HCl:
To find the pH of the solution after mixing NaOH and HCl, we need to determine the net ionic equation for the reaction between these two substances.
NaOH (base) reacts with HCl (acid) in a 1:1 molar ratio to form NaCl (salt) and H2O (water):
NaOH + HCl → NaCl + H2O
Given:
Volume of NaOH solution (V1) = 35 mL = 0.035 L
Molarity of NaOH solution (M1) = 0.23 M
Volume of HCl solution (V2) = 15 mL = 0.015 L
Molarity of HCl solution (M2) = 0.34 M
Step 1: Determine the limiting reactant (the reactant that gets completely used up).
Moles of NaOH = M1 * V1 = 0.23 M * 0.035 L = 0.00805 moles
Moles of HCl = M2 * V2 = 0.34 M * 0.015 L = 0.0051 moles
Since the mole ratio between NaOH and HCl is 1:1, HCl is the limiting reactant.
Step 2: Calculate the moles of water formed (products).
Moles of H2O = Moles of HCl (since it's a 1:1 ratio) = 0.0051 moles
Step 3: Calculate the total volume of the solution after mixing.
Total volume = Volume of NaOH + Volume of HCl = 0.035 L + 0.015 L = 0.050 L
Step 4: Calculate the concentration of OH⁻ ions and H⁺ ions in the solution after mixing.
Concentration of OH⁻ = Moles of OH⁻ / Total volume = 0.00805 moles / 0.050 L = 0.161 M
Concentration of H⁺ = Moles of H⁺ / Total volume = 0.0051 moles / 0.050 L = 0.102 M
Step 5: Calculate the pOH and pH of the solution.
pOH = -log[OH⁻] = -log(0.161) ≈ 0.79
pH = 14 - pOH = 14 - 0.79 ≈ 13.21
So, the pH of the solution after mixing 35 mL of 0.23 M NaOH with 15 mL of 0.34 M HCl is approximately 13.21.
pH of the Buffer Solution after Adding HCl:
A buffer solution is a solution that can resist changes in pH when small amounts of acid or base are added to it.
Given:
Moles of NH3 = 2.45 moles
Moles of NH4Cl = 2.05 moles
Volume of the buffer solution = 1.00 L
Step 1: Calculate the concentration of NH3 and NH4⁺ in the buffer solution.
Concentration of NH3 = Moles of NH3 / Volume of the solution = 2.45 moles / 1.00 L = 2.45 M
Concentration of NH4⁺ = Moles of NH4⁺ / Volume of the solution = 2.05 moles / 1.00 L = 2.05 M
Step 2: Calculate the initial pH of the buffer solution using the Henderson-Hasselbalch equation:
pH = pKa + log([NH3] / [NH4⁺])
Given:
Kb for NH3 = 1.8 × 10⁻⁵
Kw = Ka * Kb
1.0 × 10⁻¹⁴ = Ka * (1.8 × 10⁻⁵)
Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) ≈ 5.56 × 10⁻¹⁰
pKa = -log(Ka) ≈ -log(5.56 × 10⁻¹⁰) ≈ 9.26
Now, calculate the pH:
pH = 9.26 + log(2.45 / 2.05)
pH ≈ 9.26 + log(1.195)
pH ≈ 9.26 + 0.077
pH ≈ 9.34
Now, calculate the new pH:
pH = 9.26 + log(2.20 / 1.80)
pH ≈ 9.26 + log(1.222)
pH ≈ 9.26 + 0.088
pH ≈ 9.35
So, the pH of the solution after adding 0.25 moles of HCl to the buffer is approximately 9.35.