find the ph of asolution if 200 ml of 0.2m of NH4OH is mixed with 200ml of 0.2 HCl.given ka of HCl=1.75*10 power -5

To find the pH of the solution formed by mixing 200 mL of 0.2 M NH4OH (ammonium hydroxide) with 200 mL of 0.2 M HCl (hydrochloric acid), we need to consider the neutralization reaction between the weak base (NH4OH) and the strong acid (HCl). Let's break this down step by step.

Understanding the Reaction

When NH4OH reacts with HCl, it forms NH4Cl (ammonium chloride) and water. The reaction can be represented as:

• NH4OH + HCl → NH4Cl + H2O

Calculating Moles of Reactants

First, we need to calculate the number of moles of each reactant:

• For NH4OH:
  • Volume = 200 mL = 0.2 L
  • Concentration = 0.2 M
  • Moles of NH4OH = Volume × Concentration = 0.2 L × 0.2 mol/L = 0.04 moles
• For HCl:
  • Volume = 200 mL = 0.2 L
  • Concentration = 0.2 M
  • Moles of HCl = Volume × Concentration = 0.2 L × 0.2 mol/L = 0.04 moles

Determining the Reaction Outcome

Since we have equal moles of NH4OH and HCl (0.04 moles each), they will completely react with each other. After the reaction, we will have:

• 0 moles of NH4OH
• 0 moles of HCl
• 0.04 moles of NH4Cl in the solution

Calculating the Concentration of NH4Cl

After mixing, the total volume of the solution is:

• Total Volume = 200 mL + 200 mL = 400 mL = 0.4 L

The concentration of NH4Cl in the solution is:

• Concentration of NH4Cl = Moles / Volume = 0.04 moles / 0.4 L = 0.1 M

Finding the pH of the Solution

NH4Cl is a salt that dissociates in water to give NH4⁺ ions, which can hydrolyze to form NH4OH and H⁺ ions:

• NH4⁺ + H2O ⇌ NH4OH + H⁺

The equilibrium constant for this reaction can be derived from the ion product of water and the K_a of NH4⁺. The K_a for NH4⁺ can be calculated using the relationship:

• K_w = K_a × K_b

Given that K_w = 1.0 × 10^-14 and K_b for NH4OH is approximately 1.8 × 10^-5, we can find K_a:

• K_a = K_w / K_b = (1.0 × 10^-14) / (1.8 × 10^-5) ≈ 5.56 × 10^-10

Setting Up the Equilibrium Expression

For the hydrolysis of NH4⁺, we can set up the equilibrium expression:

• K_a = [NH4OH][H⁺] / [NH4⁺]

Assuming x is the concentration of H⁺ produced, we have:

• K_a = (x)(x) / (0.1 - x) ≈ (x²) / 0.1

Since K_a is small, we can ignore x in the denominator:

• 5.56 × 10^-10 = x² / 0.1

Solving for x gives:

• x² = 5.56 × 10^-11
• x ≈ 7.45 × 10^-6 M

Calculating the pH

The concentration of H⁺ ions is approximately 7.45 × 10^-6 M. The pH can be calculated using the formula:

• pH = -log[H⁺] = -log(7.45 × 10^-6) ≈ 5.13

Thus, the pH of the solution after mixing 200 mL of 0.2 M NH4OH with 200 mL of 0.2 M HCl is approximately 5.13.