Consider the following four titrations (i-iv): i. 150 mL of 0.2 M NH3(Kb=1.8 ×10^-5) by 0.2 M HC

Consider the following four titrations (i-iv): i. 150 mL of...

Consider the following four titrations (i-iv): i. $150 \mathrm{mL}$ of $0.2 \mathrm{M} \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)$ by $0.2 \mathrm{M} \mathrm{HCl}$ ii. $150 \mathrm{mL}$ of $0.2 \mathrm{M}$ HCl by $0.2 \mathrm{M} \mathrm{NaOH}$ iii. $150 \mathrm{mL}$ of $0.2 \mathrm{M} \mathrm{HOCl}\left(K_{\mathrm{a}}=3.5 \times 10^{-8}\right)$ by $0.2 \mathrm{M} \mathrm{NaOH}$ iv. $150 \mathrm{mL}$ of $0.2 \mathrm{M} \mathrm{HF}\left(K_{\mathrm{a}}=7.2 \times 10^{-4}\right)$ by $0.2 \mathrm{M} \mathrm{NaOH}$ a. Rank the four titrations in order of increasing $\mathrm{pH}$ at the halfway point to equivalence (lowest to highest $\mathrm{pH}$ ). b. Rank the four titrations in order of increasing $\mathrm{pH}$ at the equivalence point. c. Which titration requires the largest volume of titrant (HCl or $\mathrm{NaOH}$ ) to reach the equivalence point?

Consider the following four titrations (i-iv):
i. $150 \mathrm{mL}$ of $0.2 \mathrm{M} \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)$ by $0.2 \mathrm{M} \mathrm{HCl}$
ii. $150 \mathrm{mL}$ of $0.2 \mathrm{M}$ HCl by $0.2 \mathrm{M} \mathrm{NaOH}$
iii. $150 \mathrm{mL}$ of $0.2 \mathrm{M} \mathrm{HOCl}\left(K_{\mathrm{a}}=3.5 \times 10^{-8}\right)$ by $0.2 \mathrm{M} \mathrm{NaOH}$
iv. $150 \mathrm{mL}$ of $0.2 \mathrm{M} \mathrm{HF}\left(K_{\mathrm{a}}=7.2 \times 10^{-4}\right)$ by $0.2 \mathrm{M} \mathrm{NaOH}$
a. Rank the four titrations in order of increasing $\mathrm{pH}$ at the halfway point to equivalence (lowest to highest $\mathrm{pH}$ ).
b. Rank the four titrations in order of increasing $\mathrm{pH}$ at the equivalence point.
c. Which titration requires the largest volume of titrant (HCl or $\mathrm{NaOH}$ ) to reach the equivalence point?

Key Concepts

Stoichiometry in Titrations

Stoichiometry plays a fundamental role in titrations since the equivalence point is reached when the moles of titrant added equal the moles of analyte present. This molar balance allows for precise calculations when determining the necessary titrant volume, assuming the concentrations of the titrant and analyte are known.

Acid?Base Titration

Acid?base titration is an analytical technique where a solution of known concentration (the titrant) is gradually added to a solution of unknown concentration (the analyte) until the reaction reaches its equivalence point. This process allows the determination of the concentration of the analyte by measuring the volume of titrant required to completely neutralize the analyte.

Half?Equivalence Point and Buffer Region

The half?equivalence point in a titration of a weak acid or weak base is the stage at which half of the analyte has been neutralized. At this point, the concentrations of the weak acid and its conjugate base (or vice versa) are equal, and the pH of the solution equals the pKa (or pKb, when converted appropriately) of the analyte. This point is central to buffer behavior and is described by the Henderson–Hasselbalch equation.

Strong versus Weak Acids and Bases

Understanding the differences between strong and weak acids or bases is crucial in titration analyses. Strong acids and bases fully dissociate in water, leading to more straightforward pH calculations, while weak acids and bases only partially dissociate and require consideration of equilibrium constants (Ka or Kb) to accurately determine pH during different stages of the titration.

Equivalence Point pH and Hydrolysis

The pH at the equivalence point in a titration depends on the nature of the reactants. For titrations involving only strong acids and bases, the equivalence point pH is neutral. However, when a weak acid is titrated with a strong base (or vice versa), the conjugate base or acid formed can undergo hydrolysis, resulting in a pH that is either above or below 7. This behavior is explained by the hydrolysis of the salt formed at the equivalence point.

Transcript

00:01 This is also a titration calculation where we put together several concepts that we have learned in reference to all of the different types of titrations.

00:10 Strong acid, weak base, strong acid, strong base, weak acid, strong base, and so forth.

00:19 And the ideas associated with ph at the half equivalence point and the equivalence point.

00:28 So for the first part of this problem, it asks us to determine the ph ph at half equivalence point for all of these.

00:37 The first titration is a weak base with a strong acid, so at half equivalence we have a buffer solution where ph is going to be equal to pca.

00:49 So the pca then will be equal to the negative log of the ka value, which we can get by dividing the kb value given to us into kw, and we get a ph of 9 .26.

01:02 For the second one, it is a strong acid, strong base titration.

01:07 So at the equivalence point, we have consumed half of the hcl we started with, half the moles of hcl.

01:14 So the ph is going to be equal to the moles of hcl that is left over, which will be the moles we start with, volume times molarity, minus the moles of strong base that we added, which will be half equivalent, 75 milliliters or 0 .05 liters times its molarity and then we divide by the new volume 150 milliliters plus the 75 we added is 225 or 0 .225 liters.

01:43 This then this quotient then gives us the hydrogenium concentration we take the negative log of it to get ph and we get 1 .18.

01:52 The next one is a weak acid being titrated with a strong base so at half equivalents we have a buffer solution and ph equals pca.

02:03 And so pca will just be the negative log of the ka value, or 7 .46.

02:08 For the next one, same thing.

02:10 We just have a different ka value, and we get a pca of 3 .14.

02:15 So this one has the lowest ph at half equivalence...

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