Question

what is the ph of a solution prepared by adding 1.59 g of sodium nitrite to 195 ml of water? ka of hno2 is 4.5 × 10–4.

Answer 1

To calculate the pH of the solution, we need to determine the concentration of HNO2 and H3O+. We can use the given information and formulas to calculate these concentrations and then use the pH equation to find the pH of the solution.

Step-by-step explanation:

The pH of a solution can be calculated using the equation:

pH = -log[H3O+]

To solve this problem, we first need to determine the concentration of HNO2 in the solution. We can do this by calculating the number of moles of HNO2 using the formula:

moles = mass / molar mass

Then, we can calculate the concentration of HNO2 in mol/L using the formula:

Molarity = moles / volume (L)

Finally, we can use the concentration of HNO2 to calculate the concentration of H3O+ using the equation:

Ka = [H3O+][NO2-] / [HNO2]

Once we know the concentration of H3O+, we can use the equation pH = -log[H3O+] to determine the pH of the solution.

Answer 2

The pH of the solution can be obtained as 2.15.

The pH scale typically ranges from 0 to 14, with 7 being considered neutral. Ka, or the acid dissociation constant, is a quantitative measure of the strength of an acid in a solution. Specifically, it represents the extent to which an acid dissociates or ionizes in water.

Number of moles = 1.59 g /69 g/mol

= 0.023 moles

Molarity = 0.023 moles * 1000/195

= 0.12 M

Ka = [
`H^+`] [
`NO_3^-`]/[
`HNO_3`]

We know that;

[
`H^+`] = [
`NO_3^-`] = x

`4.5 * 10^{-4` =
`x^2`/ 0.12 - x

`4.5 * 10^{-4` ( 0.12 - x) =
`x^2`

5.4 *
`10^{-5` -
`4.5 * 10^{-4`x =
`x^2`

`x^2` +
`4.5 * 10^{-4`x - 5.4 *
`10^{-5` = 0

x = 0.007 M

pH = - log[
`H^+`]

pH = -log(0.007)

pH = 2.15