Calculating pH of a NaCN Hints & Tips

[kwikness]

Hi, I have the following problem:

Calculate the pH of a 0.200M Sodium Cyanide solution.

I'm confused because this salt has no H or O atoms in it. How could it affect the pH of a solution?
What formulas would I need to solve this? My final is tomorrow and I'm really worried. Thank you for any help.

 

[Borek]

Hint: hydrolysis. HCN is a weak acid, so CN- is a strong conjugated base. Do you remember what Bronsted-Lowry definition of acids and bases is?

 

[Blueshift5]

Hello,

First of all, don't worry! Calculating the pH of a solution containing a salt like NaCN may seem daunting, but it is actually quite straightforward.

You are correct in noting that NaCN does not contain any H or O atoms. However, it does contain the CN- ion, which is a weak base. This means that when it is dissolved in water, it will partially dissociate into CN- and Na+ ions. The CN- ion will then react with water to form OH- ions, which will affect the pH of the solution.

To solve this problem, you will need to use the dissociation constant (Kb) of CN- to calculate the concentration of OH- ions in the solution. From there, you can use the equation pH = -log[OH-] to determine the pH of the solution.

Here is the step-by-step process:

1. Write out the dissociation reaction for CN- in water: CN- + H2O ⇌ HCN + OH-

2. Write the equilibrium expression for this reaction: Kb = [HCN][OH-]/[CN-]

3. Plug in the values given in the problem: Kb = x^2/(0.200-x), where x is the concentration of OH- ions.

4. Solve for x using the quadratic formula: x = 0.00707 M.

5. Use the equation pH = -log[OH-] to calculate the pH: pH = -log(0.00707) = 2.15

So, the pH of a 0.200M NaCN solution is 2.15. This means the solution is slightly basic, as the OH- ions from the dissociation of CN- are greater in concentration than the H+ ions from the dissociation of water.

I hope this helps and good luck on your final! Remember to always double check your units and equations to ensure accurate results.