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 9.9. Find the amount of NaCN in the mixture.

To solve this problem, we can use the concept of acid-base equilibrium and the pH of the solution. The problem provides us with the total number of moles in the mixture, the volume of water, and the pH of the solution.

First, let's determine the initial concentration of the NaCN and NaHSO4 in the mixture. We will assume they completely dissociate in water.

Let's denote the amount of NaCN in moles as x, and the amount of NaHSO4 in moles as 0.60 - x (since the total is 0.60 mol).

The NaCN will undergo hydrolysis in water, reacting with water to form HCN and NaOH. This reaction goes as follows:
NaCN + H2O -> HCN + NaOH

At equilibrium, a certain amount of HCN will be formed, which will react with water to produce hydronium ions (H3O+). This reaction is an acid-base reaction:
HCN + H2O -> H3O+ + CN-

Since the pH of the solution is 9.9, we know the pOH (the negative logarithm of the hydroxide ion concentration) is 14 - 9.9 = 4.1.

To find the concentration of hydroxide ions ([OH-]), we can use the relationship between pH and pOH:
pOH = -log[OH-]

Rearranging the equation, we have:
[OH-] = 10^(-pOH)

Substituting the value, we find:
[OH-] = 10^(-4.1)

Since the solution is neutral, the concentration of hydronium ions ([H3O+]) is equal to the hydroxide ion concentration ([OH-]), which means [H3O+] = 10^(-4.1).

From the acid-base reaction, we know that the concentration of HCN is equal to the concentration of [H3O+] (because the concentration of CN- is negligible in comparison). Therefore, the concentration of HCN is also 10^(-4.1).

Now, to calculate the concentration of NaCN initially, we need to account for the hydrolysis of NaCN in water:
NaCN -> Na+ + CN-
The concentration of CN- ions is equal to the concentration of HCN formed, which is 10^(-4.1).

Since the volume of the solution is 1.0 L, the concentration of NaCN (initially) is given by:
concentration of NaCN = x mol / 1 L

Since the concentration of CN- is the same as HCN, the concentration of NaCN is:
concentration of NaCN = x mol / 1 L = 10^(-4.1) mol / 1 L

Now, we can solve for x:
x / 1 L = 10^(-4.1) mol / 1 L

Multiplying both sides by 1 L, we get:
x = 10^(-4.1) mol

So, the amount of NaCN in the mixture is approximately 10^(-4.1) mol.

To determine the amount of NaCN in the mixture, we need to use the information about the equilibrium pH and the fact that NaCN is a weak acid.

Step 1: Write the dissociation equation for NaCN.
NaCN ↔ Na+ + CN-

Step 2: Identify the acid and its conjugate base in the dissociation equation.
In this case, NaCN acts as an acid, and CN- acts as the conjugate base.

Step 3: Write the equilibrium expression (Ka) for the dissociation reaction.
Ka = [Na+][CN-] / [NaCN]

Step 4: Since NaCN is a weak acid, we can assume that the concentration of NaCN before dissociation is equal to the concentration of CN- formed after dissociation. Therefore, we can write:
[NaCN]initial = [CN-]final

Step 5: From the given information, we know that when the NaCN and NaHSO4 mixture is dissolved in 1.0 L of water, it forms a total of 0.60 mol. This means that the sum of the initial concentrations of NaCN and NaHSO4 is 0.60 mol.

So, [NaCN]initial + [NaHSO4]initial = 0.60 mol

Step 6: Let's assume that x moles of NaCN dissociate. This means that the [NaCN]initial - x = [CN-]final.

Step 7: Substitute the values obtained into the equilibrium expression (Ka).
Ka = [Na+][CN-] / [NaCN]
Ka = ([Na+])([NaCN]initial - x) / [NaCN]

Step 8: As NaCN is a weak acid, we can assume that the concentration of Na+ formed after dissociation is negligibly small compared to [NaCN]. Therefore, [NaCN]initial - x ≈ [NaCN].

Step 9: Rewrite the equilibrium expression with the approximation.
Ka = (x)([NaCN]) / [NaCN]
Ka ≈ x

Step 10: We know that pH = -log[H+], and in this case, [H+] comes from the dissociation of H2O into H+ and OH-. Since we have a pH of 9.9 (indicating a basic solution), we can assume that [OH-] is higher than [H+]. Therefore, [H+] is negligibly small compared to [OH-], and we can assume that [OH-] ≈ [CN-].

Step 11: Use the fact that pOH + pH = 14 to find the pOH of the solution.
pOH = 14 - pH
pOH ≈ 14 - 9.9
pOH ≈ 4.1

Step 12: Rewrite the pOH in terms of [OH-].
pOH = -log[OH-]
4.1 = -log[OH-]
[OH-] ≈ 10^-4.1

Step 13: Since [OH-] ≈ [CN-], we know that [CN-] ≈ 10^-4.1 M.

Step 14: Substitute the approximate value of [CN-] into the equation from Step 9.
Ka ≈ x
10^-4.1 ≈ x

Step 15: Solve for x (the amount of NaCN that dissociates).
x ≈ 10^-4.1

Step 16: The amount of NaCN in the mixture is equal to the initial concentration of NaCN minus the amount that dissociates.
[NaCN] = [NaCN]initial - x
[NaCN] ≈ [NaCN]initial - 10^-4.1

Therefore, the amount of NaCN in the mixture is approximately equal to [NaCN]initial - 10^-4.1 mol.

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