Question

Find the pH of a solution containing 0.258 g of nitrous acid in 978 mL of water

Answer 1

[AI-generated answer — Bot GPT 3.5]

To find the pH of the solution containing nitrous acid, we first need to calculate the concentration of nitrous acid in the solution.

First, calculate the molar mass of nitrous acid (HNO2):
H = 1.008 g/mol
N = 14.01 g/mol
O = 15.999 g/mol
Molar mass of HNO2 = 1.008 + 14.01 + 2(15.999) = 63.02 g/mol

Now, calculate the number of moles of nitrous acid in 0.258 g:
moles = mass / molar mass = 0.258 g / 63.02 g/mol = 0.004093 mol

Next, calculate the molarity of the nitrous acid in the solution:
Molarity (M) = moles / volume in liters
Volume in liters = 978 mL x (1 L / 1000 mL) = 0.978 L
Molarity = 0.004093 mol / 0.978 L = 0.004184 M

The dissociation of nitrous acid in water can be represented as:
HNO2 ⇌ H+ + NO2-

Since nitrous acid is a weak acid, we need to use the equilibrium constant (Ka) to find the concentration of H+ ions.
The Ka for nitrous acid is 4.5 x 10^-4.

Let x be the concentration of H+ ions, then the equilibrium concentrations will be:
[H+] = x
[HNO2] = 0.004184 - x
[NO2-] = x

Applying the expression for Ka:
Ka = [H+][NO2-] / [HNO2]
4.5 x 10^-4 = x^2 / (0.004184 - x)

Since we can assume x is small compared to 0.004184, we can simplify the expression:
4.5 x 10^-4 ≈ x^2 / 0.004184

Solving for x, we get:
x ≈ √(4.5 x 10^-4 x 0.004184) = 0.0074 M

Now that we have the concentration of H+ ions, we can calculate the pH:
pH = -log[H+]
pH = -log(0.0074) ≈ 2.13

Therefore, the pH of the solution containing 0.258 g of nitrous acid in 978 mL of water is approximately 2.13.