A 100. 0 ml sample of 0. 10 m nh3 is titrated with 0. 10 m hno3. Determine the ph of the solution after the addition of 100. 0 ml of hno3. The kb of nh3 is 1. 8 × 10-5.



Answer :

The ph of the solution after the addition of 100. 0 ml of HNO3 is 5.6. In a 100. 0 ml sample of 0. 10 m NH3 is titrated with 0. 10 m HNO3.

Define anion and cation.

Ions with a positive charge are called cations. Ions with negative charge are called anions. Ionic molecules or atoms are charged. A balanced atom will become a positively charged cation if it loses one or more electrons. An anion is a negatively charged atom that develops when a balanced atom acquires one or more electrons.

Reaction: HNO₃(aq) + NH₃(aq) → NH₄NO₃(aq) +H₂O(l)

                                                 NH₄⁺ (aq) + NO₃⁻(aq)

                NH₄⁺(aq) + H₂O (aq) ⇆ H₃O⁺(aq) + NH₃(aq)

Mol NH₃ = 0.10 mol/L × 100 mL × 1 L/1000 mL = 0.01 mol

Mol HNO₃ = 0.10 mol/L × 100 mL × 1 L/1000mL = 0.01 mol

Mol NH₄NO₃ produced = 0.01 mol NH₄NO₃

Mol HNO₃ left = 0.01 - 0.01 = 0 mol

Hydrolyzing NH₄⁺ and applying ICE approach

           NH₄⁺ --> H⁺ + NO₃⁻

I            0.01       0        0

C           -x          +x      +x

E        0.01-x       x        x

Kh = Kw/Kb = [H⁺][NO₃⁻]/[NH₄⁺]

10⁻¹⁴/1.8×10⁻⁵ = [x][x]/[0.01-x]

Solving for x,

x = [H⁺] = 2.357×10⁻⁶ mol

The formula for pH is

pH = -log [H⁺]

Aside from 2.357×10⁻⁶ mol, let's add the H⁺ from the remaining HNO₃ which is 0.

Therefore,

pH = -log[2.357×10⁻⁶ mol + 0 mol]

pH = 5.6

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