Cyanic Acid is a weak acid

HOCN + H2O <--> H3O+ +OCN- ka= 3.5*10^-4

a) if 2.5ml of 0.01 M cyanic acid solution is added to 25.0 ml of a formis acid buffer with ph = 3.70, what is the ratio of [OCN-]/[HOCN] in the resulting solution?

b) determine the pH of a 0.500 M solution of cyanic acid

c) calculate the pH of a solution that results from titrating 0.0500 moles of cyanic acid to the equivalence point with an equal number of moles of OH-, if the final volume of the solution is 150.0 mL

HOCN ==> H^+ + OCN^-

Ka = (H^+)(OCN^-)/(HOCN)
For a.
The formic acid buffer is pH 3.70 which is H^+ = 2E-4
Ka = (H^+)(OCN^-)/(HOCN)
Rearrange to
(OCN^-)/(HOCN) = Ka/(H^+)
Substitute Ka/(H^+) and you have it.

........HOCN ==> H^+ + OCN^-
initial..0.5......0......0
change.....-x.....x......x
equil....0-.5-x.....x.....x

Ka = (H^+)(OCN^-)/(HOCN)
Substitute and solve for H^+ and convert to pH.

c. The pH is determined by the hydrolysis of the salt at the equivalence point. The (salt) = 0.05mols/0.150L = ?
........OCN^- + HOH ==> HOCN + OH^-
initial..?...............0......0
change...-x..............x......x
equil....?-x..............x.....x

Kb for OCN^- = (Kw/Ka for HOCN) = (HOCN)(OH^-)/(OCN^-)
Substitute from the ICE chart and solve for x = (OH^-) and convert to pH. It will be basic.

a) Well, this question seems to be quite serious, but let me sprinkle some humor on it! Mixing cyanic acid and a formis acid buffer? Sounds like the beginning of an explosive recipe! Anyway, let's get serious now. To find the ratio of [OCN-]/[HOCN], we need to use the equation for acid dissociation: HOCN + H2O <--> H3O+ + OCN-. Since we know the Ka value is 3.5*10^-4, we can use the equation Ka = [H3O+][OCN-]/[HOCN] to find the ratio. Hope that equation doesn't give you a "Ka-boom" headache!

b) Ah, the pH of a cyanic acid solution. Well, you know what they say, "Cyanic acid might make you blue, but its pH won't make you say boo!" To find the pH, we need to use the equation for acid dissociation once again. First, calculate the concentration of H3O+ using the Ka value and then take the negative logarithm of that concentration to find the pH. It's like finding the pH of a cyanide only less deadly!

c) Titration time! Titrating cyanic acid with OH-? Sounds like a balancing act, doesn't it? Well, let's put on our pH balancing shoes and dive into it! Since we have equal moles of OH- and cyanic acid, we can assume that all the cyanic acid gets converted to OCN-. We can use the equation for acid dissociation to calculate the concentration of OCN- and then use that concentration to find the pH. Just be careful not to splash any OH- in your eyes, it might make them go "OH NO!"

a) To determine the ratio of [OCN-]/[HOCN] in the resulting solution, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([OCN-]/[HOCN])

Given that the initial pH of the buffer is 3.70, we can calculate the pOH using the equation:

pOH = 14 - pH = 14 - 3.70 = 10.30

Taking the antilog of pOH, we find the hydroxide ion concentration [OH-] in the buffer solution:

[OH-] = 10^-pOH = 10^-10.30

Since the formic acid buffer consists of a weak acid and its conjugate base, it can be assumed that [HOCN] ≈ [OCN-]. Therefore, the ratio [OCN-]/[HOCN] in the buffer solution before addition is approximately 1.

When 2.5 ml of 0.01 M cyanic acid is added to 25.0 ml of buffer solution, the volume becomes 25.0 ml + 2.5 ml = 27.5 ml. To determine the concentration of cyanic acid and the cyanate ion after dilution, we can use the dilution equation:

M1V1 = M2V2

(0.01 M)(2.5 ml) = M2(27.5 ml)

M2 = (0.01 M)(2.5 ml)/(27.5 ml) = 0.000909 M

Using the Henderson-Hasselbalch equation, we can now determine the ratio [OCN-]/[HOCN] in the resulting solution:

pH = pKa + log([OCN-]/[HOCN])

3.70 = -log(3.5*10^-4) + log([OCN-]/[HOCN])

Rearranging the equation:

log([OCN-]/[HOCN]) = 3.70 + log(3.5*10^-4)

Taking the antilog of both sides:

[OCN-]/[HOCN] = antilog(3.70 + log(3.5*10^-4))

Therefore, the ratio of [OCN-]/[HOCN] in the resulting solution is antilog(3.70 + log(3.5*10^-4)).

b) To determine the pH of a 0.500 M solution of cyanic acid, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([OCN-]/[HOCN])

Since cyanic acid is a weak acid, it does not dissociate completely. Therefore, we can assume that the cyanate ion concentration [OCN-] is approximately 0.

pH = pKa + log(0/[HOCN])

pH = pKa + log(1)

Since log(1) = 0, the pH = pKa.

Therefore, the pH of a 0.500 M solution of cyanic acid is equal to the pKa value of 3.5*10^-4.

c) To determine the pH of a solution resulting from titrating 0.0500 moles of cyanic acid to the equivalence point with an equal number of moles of OH-, we need to consider the reaction:

HOCN + OH- → OCN- + H2O

First, we need to determine the concentration of cyanic acid and OH- after the reaction has gone to completion. Since 0.0500 moles of cyanic acid are being titrated, the final volume of the solution is 150.0 mL.

The concentration of cyanic acid is:

[HOCN] = moles/volume = 0.0500 moles/ 150.0 mL

Next, we need to determine the concentration of OH- ions in the solution, which can be calculated from the volume of the solution and the number of moles of OH- added during the titration.

Since the final volume of the solution is 150.0 mL, the concentration of OH- ions is:

[OH-] = moles/volume = 0.0500 moles/150.0 mL

Finally, we can use the Henderson-Hasselbalch equation to calculate the pH of the resulting solution:

pH = pKa + log([OCN-]/[HOCN])

After the titration, we can assume that all of the cyanic acid has reacted to form the cyanate ion:

[OCN-] = moles/volume = 0.0500 moles/150.0 mL

Plugging in the values to the Henderson-Hasselbalch equation, we can calculate the pH of the resulting solution.

a) To determine the ratio of [OCN-]/[HOCN] in the resulting solution, we need to consider the buffer equation:

HOCN + H2O ⇌ H3O+ + OCN-

We are given the initial concentration of cyanic acid (HOCN) and the pH of the buffer solution. Here's how we can solve it step-by-step:

1. Convert the volume of HOCN solution to moles:
Moles of HOCN = Volume (L) × Concentration (M)
Moles of HOCN = 2.5 mL × (1 L / 1000 mL) × 0.01 M

2. Calculate the moles of OCN- at equilibrium:
Since the initial concentration of OCN- is zero, the concentration at equilibrium will be the same as the moles of HOCN reacted. Hence, moles of OCN- = moles of HOCN.

3. Use the definition of pH to calculate the concentration of H3O+ (H+):
pH = -log[H+]
[H+] = 10^(-pH)

4. Now, we can set up the equilibrium expression using the given equilibrium constant (ka):
ka = [H3O+][OCN-] / [HOCN]

5. Substitute the known values into the equation:
ka = (10^(-pH))[OCN-] / (moles of HOCN)

6. Rearrange the equation to solve for [OCN-]/[HOCN]:
[OCN-]/[HOCN] = (ka × moles of HOCN) / (10^(-pH))

Substitute the values into the equation to calculate the ratio [OCN-]/[HOCN].

b) To determine the pH of a 0.500 M solution of cyanic acid, we need to calculate the concentration of H+ in the solution. Here's how:

1. Write the ionization equation for cyanic acid:
HOCN + H2O ⇌ H3O+ + OCN-

2. Set up the equilibrium expression using the equilibrium constant (ka):
ka = [H3O+][OCN-] / [HOCN]

3. Since we are given the initial concentration of cyanic acid (HOCN) as 0.500 M, we can assume that [HOCN] at equilibrium will be approximately equal to the initial concentration.

4. Let x be the concentration (M) of H3O+ at equilibrium.

5. Substitute the known values into the equation:
ka = x^2 / 0.500

6. Solve for x by rearranging the equation and using the quadratic formula:
x = sqrt(ka * [HOCN])

7. Now, calculate the pH using the definition of pH:
pH = -log[H+]

Substitute the value of [H+] (x) into the equation to determine the pH.

c) To calculate the pH of a solution resulting from titrating 0.0500 moles of cyanic acid to the equivalence point with an equal number of moles of OH-, we need to consider the reaction between HOCN and OH-.

1. Write the balanced chemical equation for the reaction:
HOCN + OH- → OCN- + H2O

2. Determine the moles of OH- added to reach the equivalence point:
Since OH- reacts in a 1:1 mole ratio with HOCN, the moles of OH- added will be equal to the moles of HOCN.

3. Calculate the new total volume of the solution at the equivalence point. In this case, the final volume is given as 150.0 mL.

4. Calculate the concentration of OCN- at the equivalence point:
Concentration (M) = moles / volume (L)

5. Use the concentration of OCN- to calculate the pOH:
pOH = -log[OH-]

6. Finally, calculate the pH using the formula:
pH = 14 - pOH

Substitute the values into the equation to calculate the pH of the solution at the equivalence point.