The molarity of nitrous acid at which its pH becomes 2
The Correct Answer and Explanation is:
To find the molarity of nitrous acid (HNO₂) when its pH is 2, we need to use its acid dissociation constant (Ka). Nitrous acid is a weak acid with a Ka of approximately 4.5 × 10⁻⁴.
Step 1: Use the pH to find the hydrogen ion concentration
pH = 2
[H+]=10−pH=10−2=0.01 M[H⁺] = 10^{−pH} = 10^{−2} = 0.01 \, \text{M}[H+]=10−pH=10−2=0.01M
Step 2: Set up the ICE table for dissociation
Let the initial concentration of HNO₂ be C mol/L. The dissociation equation is:
HNO2⇌H++NO2−\text{HNO}_2 \rightleftharpoons \text{H}^+ + \text{NO}_2^-HNO2⇌H++NO2−
At equilibrium:
- [HNO₂] = C – x
- [H⁺] = x = 0.01
- [NO₂⁻] = x = 0.01
Step 3: Use the Ka expression
Ka=[H+][NO2−][HNO2]Ka = \frac{[H^+][NO_2^-]}{[HNO_2]}Ka=[HNO2][H+][NO2−]
Substitute known values:4.5×10−4=(0.01)(0.01)C−0.014.5 \times 10^{-4} = \frac{(0.01)(0.01)}{C – 0.01}4.5×10−4=C−0.01(0.01)(0.01)4.5×10−4=1.0×10−4C−0.014.5 \times 10^{-4} = \frac{1.0 \times 10^{-4}}{C – 0.01}4.5×10−4=C−0.011.0×10−4
Solve for C:C−0.01=1.0×10−44.5×10−4=0.222C – 0.01 = \frac{1.0 \times 10^{-4}}{4.5 \times 10^{-4}} = 0.222C−0.01=4.5×10−41.0×10−4=0.222C=0.222+0.01=0.232 MC = 0.222 + 0.01 = 0.232 \, \text{M}C=0.222+0.01=0.232M
Answer: The molarity of nitrous acid is approximately 0.232 M when the pH is 2.
Explanation )
To determine the molarity of nitrous acid (HNO₂) at which its pH becomes 2, we must understand how weak acids dissociate in water. Nitrous acid does not completely dissociate like strong acids. Instead, it partially ionizes and reaches an equilibrium. This means we can use its equilibrium constant (Ka) to find its concentration at a given pH.
The pH is defined as the negative logarithm of the hydrogen ion concentration. A pH of 2 corresponds to a hydrogen ion concentration of 0.01 M. Since nitrous acid releases one proton (H⁺) per molecule, this value also represents how much acid has dissociated into ions.
We use an ICE table, which stands for Initial, Change, and Equilibrium, to organize the concentrations. Initially, we start with some unknown concentration C of HNO₂. As it dissociates, it produces H⁺ and NO₂⁻ ions. At equilibrium, we have 0.01 M of each of these ions and (C – 0.01) M of the undissociated acid.
By substituting these into the Ka expression, we can solve for C. After rearranging and solving, we find that the molarity of nitrous acid is about 0.232 M.
This approach illustrates how weak acid behavior is tied to both pH and the acid’s Ka. It also shows that we cannot simply assume a one-to-one correspondence between concentration and pH, as we could with strong acids. Instead, we need to consider the degree of dissociation and equilibrium conditions.
