1.
Calculate the pH of a buffer solution that contains 0.32 M benzoic acid (C6H5CO2H) and 0.17 M sodium benzoate (C6H5COONa). [Ka = 6.5 × 10-5 for benzoic acid] Round your answer to two places past the decimal.
2.
A solution is prepared by mixing 470 mL of 0.18 M Tris·Base and 598 mL of 0.27 M Tris·Acid. What is the pH of this solution? [Ka(Tris·Acid) = 8.5 × 10-9] Round your answer to two places past the decimal.
3.
Calculate the pH of a buffer solution prepared by dissolving 0.21 mole of 3-chloropropanoic acid, HC3H4O2Cl and 0.86 mole of sodium 3-chloropropanoate, NaC3H4O2Cl in enough water to make 0.79 liter of solution. [Ka(HC3H4O2Cl) = 7.8 × 10-5] Round your answer to two places past the decimal.
4.
You are asked to go into the lab and prepare an acetic acid-sodium acetate buffer solution with a pH of 5.08 ± 0.02. What molar ratio of CH3COONa (conjugate base) to CH3COOH (weak acid) should be used? Read carefully. Round your answer to two places past the decimal.
5.
A buffer is prepared by adding 0.91 L of 0.99 M HCl to 815 mL of 1.5 M NaHCOO. What is the pH of this buffer? [Ka(HCOOH) = 1.7 × 10-4] Round your answer to two places past the decimal.
6.
A certain monoprotic weak acid with Ka = 0.44 can be used in various industrial processes. (a) What is the [H+] for a 0.202 M aqueous solution of this acid and (b) what is its pH? Round the [H+] to three significant figures and the pH to two places past the decimal and do NOT use scientific notation.
7.
What is the pH of 65 mL of a 1.3 M benzoic acid (HC7H5O2, Ka=6.4x10–5) solution after 20 mL of a 0.33 NaOH solution is added? Round your answer to two places past the decimal.
Why don't I do the first one and laeve the others for you. All of these are buffer problems using the Henderson-Hasselbalch equation.
1.
pH = pKa + log(base)/(acid)
pH = 6.5E-5 + log(0.17)/(0.32)= ?
4. It's the RATIO you want so solve for (base)/(acid)
1. To calculate the pH of a buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Here, pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
Given:
[HA] = 0.32 M
[A-] = 0.17 M
Ka = 6.5 × 10^-5
Substituting these values into the equation:
pH = -log(6.5 × 10^-5) + log(0.17/0.32)
Calculating the logarithms:
pH = 4.187 + log(0.5313)
Simplifying:
pH ≈ 4.19
Therefore, the pH of the buffer solution is approximately 4.19.
2. Similarly, using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Given:
[HA] = 0.18 M
[A-] = 0.27 M
Ka = 8.5 × 10^-9
Substituting these values into the equation:
pH = -log(8.5 × 10^-9) + log(0.27/0.18)
Calculating the logarithms:
pH = 8.07 + log(1.5)
Simplifying:
pH ≈ 8.44
Therefore, the pH of the solution is approximately 8.44.
3. Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Given:
[HA] = 0.21 mole
[A-] = 0.86 mole
Ka = 7.8 × 10^-5
Converting volumes to concentrations:
[HA] = 0.21 mole / 0.79 L = 0.266 M
[A-] = 0.86 mole / 0.79 L = 1.089 M
Substituting these values into the equation:
pH = -log(7.8 × 10^-5) + log(1.089/0.266)
Calculating the logarithms:
pH = 4.11 + log(4.091)
Simplifying:
pH ≈ 4.32
Therefore, the pH of the buffer solution is approximately 4.32.
4. To calculate the molar ratio of CH3COONa to CH3COOH in a buffer solution with a specific pH, we need to use the Henderson-Hasselbalch equation and rearrange it to solve for the ratio [A-]/[HA]:
pH = pKa + log([A-]/[HA])
Given:
pH = 5.08 ± 0.02 (let's assume pH = 5.08 for this calculation)
pKa = -log(Ka) = -log(1.8 x 10^-5) ≈ 4.74
Substituting these values into the equation:
5.08 ≈ 4.74 + log([A-]/[HA])
Rearranging the equation:
log([A-]/[HA]) ≈ 5.08 - 4.74
Taking the antilog of both sides:
[A-]/[HA] ≈ 10^(5.08-4.74) = 10^0.34 ≈ 2.199
Therefore, the molar ratio of CH3COONa to CH3COOH in the buffer solution should be approximately 2.199:1.
5. Let's use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Given:
[A-] = 0.91 L * 0.99 M = 0.9019 mol
[HA] = 0.815 L * 1.5 M = 1.2225 mol
Ka = 1.7 × 10^-4
Substituting these values into the equation:
pH = -log(1.7 × 10^-4) + log(0.9019/1.2225)
Calculating the logarithms:
pH = 3.77 + log(0.7377)
Simplifying:
pH ≈ 3.925
Therefore, the pH of the buffer solution is approximately 3.925.
6. (a) To calculate [H+] for a weak acid solution, we can use the equation:
[H+] = sqrt(Ka * [HA])
Given:
Ka = 0.44
[HA] = 0.202 M
Substituting these values into the equation:
[H+] = sqrt(0.44 * 0.202)
Calculating the square root:
[H+] ≈ sqrt(0.08928) ≈ 0.298
Therefore, [H+] for the 0.202 M aqueous solution of this weak acid is approximately 0.298 M.
(b) To calculate the pH, we can use the equation:
pH = -log[H+]
Substituting the value of [H+] into the equation:
pH = -log(0.298)
Calculating the logarithm:
pH ≈ 0.526
Therefore, the pH of the solution is approximately 0.53.
7. To calculate the pH after adding a base to an acid solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Given:
[HA] = 65 mL * 1.3 M = 84.5 mmol
[A-] = 20 mL * 0.33 M = 6.6 mmol
Ka = 6.4 × 10^-5
Converting volumes to concentrations:
[HA] = 84.5 mmol / 65 mL = 1.3 M
[A-] = 6.6 mmol / 85 mL = 0.0776 M
Substituting these values into the equation:
pH = -log(6.4 × 10^-5) + log(0.0776/1.3)
Calculating the logarithms:
pH = 4.194 + log(0.0597)
Simplifying:
pH ≈ 4.29
Therefore, the pH of the solution after adding NaOH is approximately 4.29.