The following questions refer to the following system:
A total of 30.0 mL of a .10M solution of a monophonic acid (Ka = 1.0 x 10 ^-5) is titrated with .20M sodium hydroxide solution.
1) Before the titration begins, the pH of the solution is about
A) 2
B) 5
C) 7
D) 9
E) 12
2) Af the equivalence point, the pH of the solution is about
A) 2
B) 5
C) 7
D) 9
E) 12
3) The amount of NaOH required to reach the equivalence point is
A) 15.0 mL
B) 30.0 mL
C) 45.0 mL
D) 50.0 mL
E) 60.0 mL
4) The approximate pH of the solution when the weak acid is half neutralized is
A) 2
B) 5
C) 7
D) 9
E) 12
5) What is true of a buffer made from the weak acid that has a pH of 4.0?
A) [HA] = [A-]
B) [HA] = 10[A-]
C) 10[HA] = [A-]
D) 2[HA] = [A-]
E) [HA] = 2[A-]
Explanations would be appreciated
I strongly suspect you know how to do some of these and for those it's redundant for me to help. Post your work on what you know and clearly explain what you don't understand about the others and I'll help you through them.
1) To determine the pH of the solution before the titration begins, we need to consider the dissociation of the weak acid. The given Ka value (acid dissociation constant) is 1.0 x 10^-5. Since the concentration of the weak acid is 0.10 M, we can assume that the amount of H+ ions produced by the dissociation of the acid is negligible compared to the concentration of H+ ions present in water.
Therefore, the pH of the solution before the titration begins is approximately equal to the pH of pure water, which is 7. So, the answer is C) 7.
2) At the equivalence point of a titration, the moles of acid and base are stoichiometrically equal, resulting in the formation of a salt and water. In this case, the weak acid is completely neutralized by the base, resulting in the formation of the salt of the acid.
Since the salt does not undergo any significant hydrolysis, the pH at the equivalence point will be determined by the level of hydroxide ions present in the solution. Considering that sodium hydroxide is a strong base, its complete ionization produces OH- ions.
The concentration of OH- ions in the solution can be calculated using the equation M1V1 = M2V2, where M1 is the concentration of the base, V1 is the volume of the base used at the equivalence point, M2 is the concentration of a resulting solution, and V2 is its volume.
Given that the concentration of the base is 0.20 M and the volume is 30.0 mL, and assuming that the volumes of the acid and base solutions are equal at the equivalence point, we can calculate the concentration of the salt solution produced.
M1V1 = M2V2
(0.20 M)(30.0 mL) = (M2)(60.0 mL)
Solving for M2:
M2 = (0.20 M)(30.0 mL) / (60.0 mL)
M2 = 0.10 M
Since the concentration of OH- ions is equal to the concentration of the resulting salt solution, we can use the equation pOH = -log[OH-] to find the pOH. Then, we can convert the pOH to pH using the equation pH + pOH = 14.
pOH = -log[OH-]
pOH = -log(0.10)
pOH ≈ 1
pH + pOH = 14
pH + 1 = 14
pH ≈ 13
Therefore, the pH of the solution at the equivalence point is approximately 13. So, the answer is E) 12.
3) To determine the amount of NaOH required to reach the equivalence point, we need to consider the stoichiometry of the reaction between the acid and the base. The balanced chemical equation for this neutralization reaction is:
Weak acid (HA) + Sodium hydroxide (NaOH) → Sodium salt of the acid (NaA) + Water (H2O)
From the equation, we can see that one mole of the weak acid reacts with one mole of sodium hydroxide at the equivalence point.
The number of moles of the weak acid can be calculated using the formula: moles = concentration x volume. Since the concentration is 0.10 M and the volume is 30.0 mL, we have:
Moles of weak acid = (0.10 M)(30.0 mL) = 3.0 mmol
Since the stoichiometric ratio of the reaction is 1:1, the number of moles of sodium hydroxide required to neutralize the weak acid is also 3.0 mmol.
To convert the moles of sodium hydroxide to volume, we use the equation moles = concentration x volume. The concentration of the sodium hydroxide is 0.20 M, and we want to find the volume:
3.0 mmol = (0.20 M)(volume)
volume = 3.0 mmol / 0.20 M
volume = 15.0 mL
Therefore, the amount of NaOH required to reach the equivalence point is 15.0 mL. So, the answer is A) 15.0 mL.
4) When the weak acid is half neutralized, it means that half of the moles of the weak acid have reacted with the sodium hydroxide. Since the reaction between a weak acid and a strong base is stoichiometric (1:1 in this case), we can assume that half of the original concentration of the weak acid has been consumed.
The initial concentration of the weak acid is 0.10 M and the volume is 30.0 mL. Therefore, half neutralization occurs when the moles of weak acid remaining is equal to half of the original moles:
Moles of weak acid remaining = (0.10 M)(30.0 mL) / 2 = 1.5 mmol
To find the concentration of the remaining weak acid, we divide the remaining moles by the volume:
Concentration of weak acid remaining = 1.5 mmol / 30.0 mL = 0.050 M
Since the weak acid is only partially neutralized, we can consider it as a weak acid with a new concentration. To find the pH of this remaining weak acid solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
The pKa value given for the weak acid is 1.0 x 10^-5. Since the weak acid is half neutralized, the concentration of the weak acid ([HA]) will be half of its initial concentration.
pH = -log(1.0 x 10^-5) + log (0.050 / 0.100)
pH ≈ 5 + log (0.500)
pH ≈ 5 - 0.30
pH ≈ 4.7
Therefore, the approximate pH of the solution when the weak acid is half neutralized is 4. So, the answer is B) 5.
5) A buffer solution is formed by the combination of a weak acid and its conjugate base (salt). A buffer solution resists changes in pH when small amounts of acid or base are added.
For a buffer to be effective, the concentrations of the weak acid (HA) and its conjugate base (A-) should be approximately equal. In other words, the ratio of [HA] to [A-] should be close to 1:1.
Given that the pH of the weak acid is 4.0, we can assume that the weak acid is partially dissolved and dissociated. Using the Henderson-Hasselbalch equation, we have:
pH = pKa + log ([A-] / [HA])
Since we have a pH of 4.0, we can rearrange the equation to solve for the ratio [A-] / [HA]:
4.0 = -log (Ka) + log ([A-] / [HA])
4.0 + log (Ka) = log ([A-] / [HA])
Using the given Ka value of 1.0 x 10^-5, we calculate:
4.0 + log (1.0 x 10^-5) = log ([A-] / [HA])
log ([A-] / [HA]) ≈ -9
We can approximate that the ratio of [A-] to [HA] is approximately 1:10^9. Since the pH of the weak acid is lower than the pKa value, the concentration of [HA] is higher than the concentration of [A-].
Therefore, the correct statement for the buffer made from the weak acid with a pH of 4.0 is D) 2[HA] = [A-].