calculate the ph of ( 1.5\times10^{-4}m nano_{2} ). (the ( k_{a} ) for ( hno_{2} ) is ( 4.0\times10^{-4} ).)

calculate the ph of ( 1.5\times10^{-4}m nano_{2} ). (the ( k_{a} ) for ( hno_{2} ) is ( 4.0\times10^{-4} ).)

calculate the ph of ( 1.5\times10^{-4}m nano_{2} ). (the ( k_{a} ) for ( hno_{2} ) is ( 4.0\times10^{-4} ).)

Answer

Explanation:

Step1: Find $K_b$ of $\text{NO}_2^-$

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{4.0 \times 10^{-4}} = 2.5 \times 10^{-11}$

Step2: Set up hydrolysis equation

$\text{NO}_2^- + \text{H}_2\text{O} \rightleftharpoons \text{HNO}_2 + \text{OH}^-$
Let $[\text{OH}^-] = x$, then $K_b = \frac{x^2}{c - x} \approx \frac{x^2}{c}$ (since $K_b$ is small)

Step3: Calculate $x$

$x = \sqrt{K_b \times c} = \sqrt{2.5 \times 10^{-11} \times 1.5 \times 10^{-4}} = \sqrt{3.75 \times 10^{-15}} \approx 6.12 \times 10^{-8}$

Step4: Find $\text{pOH}$ then $\text{pH}$

$\text{pOH} = -\log(6.12 \times 10^{-8}) \approx 7.21$
$\text{pH} = 14 - 7.21 = 6.79$

Answer:

6.79