Question

use the references to access important values if needed for this question. the ph of an aqueous solution of 0.120 m sodium nitrite, nano₂, is (assume that kₐ(hno₂) = 4.50 × 10⁻⁴). this solution is

Explanation:

Step1: Calculate \(K_b\) for \(NO_2^-\)

We know that \(K_w=K_a\times K_b\), where \(K_w = 1.0\times10^{-14}\). Given \(K_a(HNO_2)=4.50\times 10^{-4}\), then \(K_b=\frac{K_w}{K_a}=\frac{1.0\times 10^{-14}}{4.50\times 10^{-4}}\)

$$K_b=\frac{1.0\times 10^{-14}}{4.50\times 10^{-4}}=\frac{1}{4.50}\times10^{-10}\approx2.22\times 10^{-11}$$

Step2: Set up the hydrolysis equation and equilibrium expression

The hydrolysis of \(NO_2^-\) is \(NO_{2}^{-}+H_{2}O
ightleftharpoons HNO_{2}+OH^{-}\). Let \(x\) be the concentration of \(OH^{-}\) and \(HNO_2\) at equilibrium. The initial concentration of \(NO_2^-\) is \(c = 0.120\space M\). The equilibrium expression is \(K_b=\frac{[HNO_{2}][OH^{-}]}{[NO_{2}^{-}]}\). Assuming \(x\) is small compared to \(0.120\space M\) (since \(K_b\) is very small), \([NO_{2}^{-}]\approx0.120\space M\), and \([HNO_{2}]=[OH^{-}]=x\). So \(K_b=\frac{x\times x}{0.120}\)

Step3: Solve for \(x\) (\([OH^{-}]\))

Substitute \(K_b = 2.22\times 10^{-11}\) into \(K_b=\frac{x^{2}}{0.120}\). Then \(x^{2}=K_b\times0.120\)

$$x^{2}=2.22\times 10^{-11}\times0.120 = 2.664\times 10^{-12}$$

$$x=\sqrt{2.664\times 10^{-12}}\approx1.63\times 10^{-6}\space M$$

Step4: Calculate \(pOH\)

$$pOH=-\log([OH^{-}])=-\log(1.63\times 10^{-6})$$

$$pOH = 5.79$$

Step5: Calculate \(pH\)

Since \(pH + pOH=14\), then \(pH=14 - pOH\)

$$pH=14 - 5.79=8.21$$

Answer:

\(8.21\)