Calculate the pH of the the solution that results from titrating 50 mL of 0.6 M HCN(aq) with 0.2 M NaOH

At 0.00 mL of NaOH
At 85 mL of NaOH.
At 150 mL NaOH.
At 160 mL of NaOH

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To determine the pH at different points during the titration, we need to calculate the moles of acid and base present and then use the appropriate equilibrium expression to find the concentration of the hydronium ion (H3O+). The pH can be calculated using the equation pH = -log[H3O+].

Given:
Initial volume of HCN (V1) = 50 mL
Concentration of HCN (C1) = 0.6 M
Volume of NaOH (V2) added = 0 mL, 85 mL, 150 mL, 160 mL
Concentration of NaOH (C2) = 0.2 M

1. At 0.00 mL of NaOH:
The moles of HCN initially present can be calculated as:
moles HCN = volume (V1) x concentration (C1)
moles HCN = 50 mL x 0.6 M
moles HCN = 30 mmol

Since no NaOH has been added yet, there are no moles of base present.

To determine the pH, we need to calculate the concentration of hydronium ions (H3O+):
HCN + H2O ⇌ H3O+ + CN-

The initial concentration of HCN is equal to the concentration of hydronium ions at this point.

[H3O+] = [HCN initial] = moles HCN / total volume (V1)
[H3O+] = 30 mmol / (50 mL)
[H3O+] = 0.6 M

pH = -log(0.6)
pH = 0.22

So, at 0.00 mL of NaOH, the pH is approximately 0.22.

2. At 85 mL of NaOH:
To determine the moles of base added, we use the equation:
moles base = volume (V2) x concentration (C2)
moles base = 85 mL x 0.2 M
moles base = 17 mmol

Since HCN and NaOH react in a 1:1 ratio, the moles of HCN remaining after the reaction can be calculated as:
moles HCN remaining = moles HCN initial - moles base added
moles HCN remaining = 30 mmol - 17 mmol
moles HCN remaining = 13 mmol

To determine the concentration of hydronium ions, we use the Henderson-Hasselbalch equation for weak acids:
pH = pKa + log ([A-]/[HA])

At the half-equivalence point (where moles base added = moles initial acid / 2), the weak acid and its conjugate base are present in equal amounts:
[H3O+] = [CN-]

pKa for HCN is given as 9.24.

pH = 9.24 + log ([CN-] / [HCN remaining])
pH = 9.24 + log(1)
pH = 9.24

So, at 85 mL of NaOH, the pH is approximately 9.24.

3. At 150 mL of NaOH:
Moles base added = 150 mL x 0.2 M = 30 mmol

Moles HCN remaining = 30 mmol - 30 mmol = 0 mmol

Since all the HCN has reacted, we have an excess of base. The solution will be basic, and the pH will be determined by the concentration of hydroxide ions (OH-).

[OH-] = moles base added / total volume (V1 + V2)
[OH-] = 30 mmol / (50 mL + 150 mL)
[OH-] = 0.15 M

pOH = -log(0.15)
pOH = 0.82

pH = 14 - pOH
pH = 14 - 0.82
pH = 13.18

So, at 150 mL of NaOH, the pH is approximately 13.18.

4. At 160 mL of NaOH:
Moles base added = 160 mL x 0.2 M = 32 mmol

Moles HCN remaining = 30 mmol - 32 mmol = -2 mmol

Since the moles of base added exceed the initial moles of acid, the excess base will determine the pH.

[OH-] = moles base added / total volume (V1 + V2)
[OH-] = 32 mmol / (50 mL + 160 mL)
[OH-] = 0.14 M

pOH = -log(0.14)
pOH = 0.85

pH = 14 - pOH
pH = 14 - 0.85
pH = 13.15

So, at 160 mL of NaOH, the pH is approximately 13.15.

To calculate the pH of the solution resulting from titration, we need to determine the moles of acid (HCN) and the moles of base (NaOH) at each point in the titration. Then, we can use the balanced chemical equation and the concept of stoichiometry to determine the amount of the remaining acid and base.

First, let's calculate the moles of acid and base at each point:

1. At 0.00 mL of NaOH:
- Moles of HCN = (volume of HCN x molarity of HCN) / 1000 = (50 mL x 0.6 M) / 1000 = 0.03 moles
- Moles of NaOH = 0.00 moles (as no NaOH is added yet)

2. At 85 mL of NaOH:
- Moles of HCN = 0.03 moles (initial amount)
- Moles of NaOH = (volume of NaOH x molarity of NaOH) / 1000 = (85 mL x 0.2 M) / 1000 = 0.017 moles

3. At 150 mL of NaOH:
- Moles of HCN = 0.03 moles - moles of NaOH added (from previous step)
- Moles of NaOH = (volume of NaOH x molarity of NaOH) / 1000 = (150 mL x 0.2 M) / 1000 = 0.03 moles

4. At 160 mL of NaOH:
- Moles of HCN = 0.03 moles - moles of NaOH added (from previous step)
- Moles of NaOH = (volume of NaOH x molarity of NaOH) / 1000 = (160 mL x 0.2 M) / 1000 = 0.032 moles

Now, let's calculate the pH at each point using the Henderson-Hasselbalch equation, which relates the pH to the concentration of the acid and base:

pH = pKa + log ([A-] / [HA])

In this case, the acid is HCN (hydrocyanic acid), and the base is NaCN (sodium cyanide). The pKa of HCN is 9.21, which is the negative logarithm of the acid dissociation constant.

1. At 0.00 mL of NaOH:
- Concentration of [A-] = moles of CN- / total volume (HCN + NaCN)
- Concentration of [HA] = moles of HCN / total volume
- pH = 9.21 + log ([CN-] / [HCN])

2. At 85 mL of NaOH:
- Concentration of [A-] = moles of CN- / total volume (HCN + NaCN)
- Concentration of [HA] = moles of HCN / total volume
- pH = 9.21 + log ([CN-] / [HCN])

3. At 150 mL of NaOH:
- Concentration of [A-] = moles of CN- / total volume (HCN + NaCN)
- Concentration of [HA] = moles of HCN / total volume
- pH = 9.21 + log ([CN-] / [HCN])

4. At 160 mL of NaOH:
- Concentration of [A-] = moles of CN- / total volume (HCN + NaCN)
- Concentration of [HA] = moles of HCN / total volume
- pH = 9.21 + log ([CN-] / [HCN])

By substituting the values appropriately, you can calculate the pH at each point in the titration.