What is the ph of a solution containing .12mol/l nh4cl and .03mol/l naoh?

Answers

Answer 1

To determine the pH of the solution, we first need to calculate the concentration of the resulting solution after the reaction between NH4Cl and NaOH.

The balanced chemical equation for the reaction is:

NH4Cl + NaOH → NaCl + NH3 + H2O

From the equation, we can see that NH4Cl reacts with NaOH to form NaCl, NH3, and H2O.

The NH3 produced will react with water to form NH4+ and OH- ions. Therefore, the resulting solution will contain NH4+, Cl-, Na+, and OH- ions.

To calculate the concentration of NH4+ and OH- ions, we need to use the following equations:

[tex]NH4Cl → NH4+ + Cl-[/tex]

[tex]NaOH → Na+ + OH-[/tex]

The NH4+ and OH- ions will react according to the following equation:

[tex]NH4+ + OH- → NH3 + H2O[/tex]

We can use the initial concentrations of NH4Cl and NaOH to calculate the concentration of NH4+ and OH- ions in the resulting solution:

[ NH4+ ] = 0.12 mol/L

[ OH- ] = 0.03 mol/L

To calculate the pH, we need to determine the concentration of H+ ions in the solution. Since NH4+ is a weak acid, it will undergo partial dissociation according to the following equation:

[tex]NH4+ + H2O ↔ NH3 + H3O+[/tex]

The equilibrium constant expression for this reaction is:

Ka = [ NH3 ][ H3O+ ] / [ NH4+ ]

Since NH4+ is the limiting reactant, we can assume that all of the NH4+ ions will react to form NH3 and H3O+ ions. Therefore, the concentration of NH3 and H3O+ ions will be equal to [ NH4+ ].

[ NH3 ] = [ NH4+ ] = 0.12 mol/L

Substituting the values into the equilibrium constant expression and solving for [ H3O+ ], we get:

[tex]Ka = 5.6 × 10^-10[/tex]

[tex][ H3O+ ] = sqrt( Ka × [ NH4+ ] ) = 1.34 × 10^-6 mol/L[/tex]

pH = -log [ H3O+ ] = -log ( 1.34 × 10^-6 ) = 5.87

Therefore, the pH of the solution is 5.87.

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