Question

a solution is made by dissolving 35.5 g of ba(no₂)₂ in 500.0 ml of water. using kb(no₂⁻) = 2.2 × 10⁻¹¹, determine the ph of the solution.

Explanation:

Step1: Calculate the molarity of \(Ba(NO_2)_2\)

The molar mass of \(Ba(NO_2)_2\) is \(M = 137.3 + 2\times(14 + 2\times16)= 229.3\space g/mol\).
The number of moles of \(Ba(NO_2)_2\), \(n=\frac{m}{M}=\frac{35.5\space g}{229.3\space g/mol}\approx0.155\space mol\).
The volume of the solution \(V = 500.0\space mL=0.500\space L\).
The molarity of \(Ba(NO_2)_2\), \(C=\frac{n}{V}=\frac{0.155\space mol}{0.500\space L}=0.31\space M\).
Since \(Ba(NO_2)_2 = Ba^{2 +}+2NO_2^{-}\), the molarity of \(NO_2^{-}\) is \(C(NO_2^{-}) = 2\times0.31\space M = 0.62\space M\).

Step2: Set up the hydrolysis equation and \(K_b\) expression

The hydrolysis of \(NO_2^{-}\) is \(NO_2^{-}+H_2O
ightleftharpoons HNO_2 + OH^{-}\).
Let \(x\) be the concentration of \(OH^{-}\) and \(HNO_2\) at equilibrium. Then the concentration of \(NO_2^{-}\) at equilibrium is \((0.62 - x)\space M\).
\(K_b=\frac{[HNO_2][OH^{-}]}{[NO_2^{-}]}\), and since \(K_b = 2.2\times 10^{-11}\) is very small, \(0.62-x\approx0.62\).
So \(2.2\times 10^{-11}=\frac{x\cdot x}{0.62}\).

Step3: Solve for \(x\) (concentration of \(OH^{-}\))

\(x^{2}=2.2\times 10^{-11}\times0.62\), \(x^{2}=1.364\times 10^{-11}\), \(x=\sqrt{1.364\times 10^{-11}}\approx3.69\times 10^{-6}\space M=[OH^{-}]\).

Step4: Calculate \(pOH\) and \(pH\)

\(pOH=-\log[OH^{-}]=-\log(3.69\times 10^{-6})\approx5.43\).
Since \(pH + pOH=14\), \(pH = 14 - 5.43=8.57\).

Answer:

\(8.57\)