Calculate Oh and Ph for 0.10 M Nacn
This calculator helps determine the hydroxide ion concentration ([OH-]) and pH of a 0.10 M sodium cyanide (NaCN) solution. Sodium cyanide is a weak electrolyte that dissociates in water, forming cyanide ions (CN-) and sodium ions (Na+). The dissociation of HCN (a weak acid) affects the pH of the solution.
Introduction
When sodium cyanide dissolves in water, it partially dissociates into HCN and CN- ions. The HCN then dissociates into H+ and CN- ions, which affects the pH of the solution. This calculator uses the dissociation constant of HCN to determine the [OH-] and pH.
Key Assumptions:
- The dissociation constant of HCN (Ka) is 4.9 × 10-10 at 25°C.
- The solution is dilute, so activity coefficients are assumed to be 1.
- The autoionization of water is considered (Kw = 1.0 × 10-14).
Formula
The pH of the solution is calculated using the following steps:
- Determine the concentration of HCN formed from NaCN dissociation.
- Calculate the concentration of H+ ions using the dissociation constant of HCN.
- Determine the concentration of OH- ions using the autoionization of water.
- Calculate the pH from the H+ concentration.
Ka = [H+][CN-]/[HCN]
[H+] = √(Ka × [HCN])
[OH-] = Kw/[H+]
pH = -log[H+]
Calculation
The calculation involves solving a quadratic equation derived from the dissociation of HCN. The exact steps are implemented in the calculator below.
Example
For a 0.10 M NaCN solution:
- First, determine the concentration of HCN: [HCN] = 0.10 M × 0.01 = 0.001 M (assuming 1% dissociation).
- Calculate [H+] using Ka: [H+] = √(4.9 × 10-10 × 0.001) ≈ 7.0 × 10-6 M.
- Determine [OH-]: [OH-] = 1.0 × 10-14/7.0 × 10-6 ≈ 1.4 × 10-9 M.
- Calculate pH: pH = -log(7.0 × 10-6) ≈ 5.15.
FAQ
What is the dissociation constant of HCN?
The dissociation constant (Ka) of HCN is approximately 4.9 × 10-10 at 25°C.
Why does NaCN affect pH?
NaCN dissociates into HCN and CN-, and HCN further dissociates into H+ and CN-, which lowers the pH.
Is the autoionization of water considered?
Yes, the calculator uses Kw = 1.0 × 10-14 to determine [OH-].