Calculate hydroxyl ion concentration, ph and the percent hydrolysis for 0.1M solution of sodium cyanide (hints know for cyanide ion is 2.5*10^-5).
I don't understand the hint. What's know. I assume that is some K for cyanide and my best guess is that is Ka for HCN. Here is how you do it knowing Ka.
..............CN^- HOH ==>HCN + OH^-
I.............0.1........................0..........0
C.............-x........................x...........x
E,,,,.......0.1-x......................x...........x
Kb for CN^- = (Kw/Ka for HCN) = (x)(x)/(0.1-x)
Substitute and solve for x = (OH^-). Convert that to pH .
%hydrolysis = [(OH^-)/0.1]*100
To calculate the hydroxyl ion concentration, pH, and percent hydrolysis for a 0.1M solution of sodium cyanide, we need to determine the concentration of hydroxyl ions, which is related to the hydrolysis of cyanide ions in water.
Here are the step-by-step calculations:
Step 1: Write the balanced chemical equation for the hydrolysis of sodium cyanide:
CN- + H2O ↔ HCN + OH-
Step 2: Set up an equilibrium expression for the hydrolysis reaction:
Kw = [HCN] [OH-] / [CN-]
Step 3: Substitute the known values:
Kw = x * (0.1 - x) / (0.1)
Since the concentration of HCN is negligible compared to the initial concentration of CN- (0.1M), the concentration of hydroxyl ions [OH-] is equal to x.
Step 4: Use the hydrolysis constant of cyanide ion (2.5 x 10^-5):
Kw = (2.5 x 10^-5) * (0.1 - 2.5 x 10^-5) / (0.1)
1.0 x 10^-14 = (2.5 x 10^-5) * (0.1 - 2.5 x 10^-5) / (0.1)
Step 5: Solve the equation for x (concentration of hydroxyl ions):
x = (1.0 x 10^-14) * (0.1) / (2.5 x 10^-5 - 0.1)
x ≈ 4 x 10^-9 M (concentration of hydroxyl ions)
Step 6: Calculate the pH:
pOH = -log10 [OH-]
pOH = -log10(4 x 10^-9)
pOH ≈ 8.4
pH = 14 - pOH
pH ≈ 14 - 8.4
pH ≈ 5.6
Step 7: Calculate the percent hydrolysis:
% Hydrolysis = [OH-] / (initial concentration of CN-) * 100
% Hydrolysis = (4 x 10^-9) / (0.1) * 100
% Hydrolysis ≈ 4 x 10^-6 %
Therefore, for a 0.1M solution of sodium cyanide, the hydroxyl ion concentration is approximately 4 x 10^-9 M, the pH is approximately 5.6, and the percent hydrolysis is approximately 4 x 10^-6%.
To calculate the hydroxyl ion concentration, pH, and percent hydrolysis of a 0.1M solution of sodium cyanide (NaCN), you need to consider the cyanide ion (CN-) concentration and its behavior in water.
Step 1: Determine the cyanide ion concentration:
The problem states that the concentration of cyanide ion (CN-) is 2.5*10^-5. This value represents the concentration of CN- in the sodium cyanide solution because sodium (Na+) does not significantly react with water.
Step 2: Write the reaction equation for hydrolysis:
Cyanide ion (CN-) reacts with water (H2O), resulting in the hydroxyl ion (OH-) and a weak acid called hydrocyanic acid (HCN).
CN- + H2O ⇌ OH- + HCN
Step 3: Using the cyanide ion concentration, calculate the hydroxyl ion concentration:
Since the concentration of CN- is given, it is the same as the concentration of OH- produced by hydrolysis.
[OH-] = 2.5*10^-5 M
Step 4: Calculate the pH:
The pH can be determined using the hydroxyl ion concentration ([OH-]). Since [OH-] is given, we can use the following equation to calculate the pH:
pOH = - log10 [OH-]
pH = 14 - pOH
Let's calculate pOH first:
pOH = - log10 (2.5*10^-5)
pOH ≈ 4.60
Now, calculate the pH:
pH = 14 - 4.60
pH ≈ 9.40
Therefore, the pH of the solution is approximately 9.40.
Step 5: Calculate percent hydrolysis:
The percent hydrolysis represents the amount of CN- that has undergone hydrolysis. To calculate it, we need to compare the hydroxyl ion concentration ([OH-]) from the hydrolysis reaction with the original CN- concentration.
Percent hydrolysis = ([OH-]/[CN-]) * 100
Substituting the values:
Percent hydrolysis = (2.5*10^-5 M / 0.1 M) * 100
Percent hydrolysis ≈ 0.025%
Therefore, the percent hydrolysis of the 0.1M sodium cyanide solution is approximately 0.025%.
Remember, when solving similar problems, always double-check the assumptions made and units used to avoid any mistakes in the calculations.