A mixture of NaCN and NaHSO4 consists of a total of 0.60 mol. When the mixture is dissolved in 1.0 L of water and comes to equilibrium the pH is found to be 9.7.
To find the pH of the solution, we need to understand the components of the mixture and their behavior in water.
1. Identify the components: The given mixture consists of NaCN (sodium cyanide) and NaHSO4 (sodium bisulfate).
2. Understand the behavior of the components in water:
- NaCN dissociates completely into Na+ and CN- ions.
- NaHSO4 dissociates into Na+ and HSO4- ions.
- HSO4- can partially dissociate further into H+ and SO4^2- ions.
3. Identify what contributes to the pH of the solution:
- The H+ ions from the partial dissociation of HSO4- contribute to the acidity of the solution.
- The OH- ions from the water dissociation contribute to the basicity of the solution.
Knowing this, let's calculate the concentrations of H+ and OH- ions to determine the pH using the equilibrium constants.
4. Set up the relevant equations and expressions:
- The dissociation of HSO4-:
HSO4- ⇌ H+ + SO4^2-
- The dissociation of water:
H2O ⇌ H+ + OH-
- The equilibrium expression for water:
Kw = [H+][OH-] = 1.0 x 10^-14 (at 25°C)
5. Calculate the concentration of OH- ions:
From the equilibrium expression for water, we can find the concentration of OH- ions.
[OH-] = Kw / [H+]
6. Calculate the concentration of H+ ions:
Since the concentration of OH- ions is related to the concentration of H+ ions by the equilibrium constant, we can substitute the previous result into the equation.
[H+] * [OH-] = Kw
[H+] = Kw / [OH-]
7. Find the concentration of H+ ions:
Now, we need to find [OH-] using the known pH value of 9.7.
pH = -log[H+]
[H+] = 10^(-pH)
8. Calculate [OH-]:
Substitute the [H+] value from the previous step into the equation [OH-] = Kw / [H+].
9. Calculate pOH:
pOH = -log[OH-]
10. Calculate pH:
pH = 14 - pOH
By following these steps, the pH of the solution can be calculated.