Question

Ammonia is a weak base with a K_bof 1.8$\times$10^-6. A 140mL sample of a 0.175M solution of aqueous ammonia is titrated with a 0.106M solution of the strong acid HCl . The reaction is

NH₃(aq) + HCl (aq)$\to$NH₄^-(aq) +Cl^- (aq)

Short answer

pH of the solution after the addition of base is 11.25 .

Step-by-step solution

Identifying the Logarithm 

The negative logarithm of hydrogen ion concentration is called as pH .

Mathematically, pH = -log{H}

The negative logarithm of the hydroxide (OH^-) ion concentration of a solution is called as pOH.

pOH = -log{OH^-}

The values pH and pOH are related to each other as shown below:

pK_b=pH + pOH

Calculating pH

Calculate the pH of NH₃ before addition of acid to it by applying the formula below. Since NH₃ is a weak base, let the change in equilibrium concentration of base bey mol / L . The value of K_b is 1.8$\times$10^-5.

Calculating the value if y

Substitute the following values for the equilibrium constant in the equation:

K_{b=$\frac{\left[N{H}_4\right]\left[O{H}^{-}\right]}{\left[N{H}_3\right]}$}

1.8$\times$10^{-5=$\frac{y^2}{(0.175-y)}$}

Since K_b is a small value, we can assume that y<<0.175 . Thus, we can ignore y in the denominator and modify the equation.

  Calculating pOH

1.8$\times$10^-5=$\frac{y^2}{0.175-y}$

y = 1.77$\times$10^-3

[OH^-] = 1.77$\times$10^-3

Let us calculate the pOH of the solution by applying the following formula

pOH = -log[OH^-]

= -log[ 1.77$\times$10^-3]

pOH = 2.752

Now calculate the pH by applying the formula below.

pK_b= pH + pOH