Construct a rough plot of pH versus volume of base for the titration of 25.0 mL of 0.050 M HCN

Construct a rough plot of pH versus volume of base for the...

Construct a rough plot of $\mathrm{pH}$ versus volume of base for the titration of $25.0 \mathrm{~mL}$ of $0.050 \mathrm{M} \mathrm{HCN}$ with $0.075 \mathrm{M} \mathrm{NaOH}$. (a) What is the $\mathrm{pH}$ before any $\mathrm{NaOH}$ is added? (b) What is the $\mathrm{pH}$ at the halfway point of the titration? (c) What is the $\mathrm{pH}$ when $95 \%$ of the required $\mathrm{NaOH}$ has been added? (d) What volume of base, in milliliters, is required to reach the equivalence point? (e) What is the $\mathrm{pH}$ at the equivalence point? (f) What indicator would be most suitable for this titration? (Figure 17.11.) (g) What is the $\mathrm{pH}$ when $105 \%$ of the required base has been added?

Construct a rough plot of $\mathrm{pH}$ versus volume of base for the titration of $25.0 \mathrm{~mL}$ of $0.050 \mathrm{M} \mathrm{HCN}$ with $0.075 \mathrm{M} \mathrm{NaOH}$.
(a) What is the $\mathrm{pH}$ before any $\mathrm{NaOH}$ is added?
(b) What is the $\mathrm{pH}$ at the halfway point of the titration?
(c) What is the $\mathrm{pH}$ when $95 \%$ of the required $\mathrm{NaOH}$ has been added?
(d) What volume of base, in milliliters, is required to reach the equivalence point?
(e) What is the $\mathrm{pH}$ at the equivalence point?
(f) What indicator would be most suitable for this titration? (Figure 17.11.)
(g) What is the $\mathrm{pH}$ when $105 \%$ of the required base has been added?

Key Concepts

Indicator Selection

Indicator selection is a crucial aspect of titration that ensures the visible endpoint closely matches the true equivalence point. The chosen indicator should have a pH transition range that falls within the steep part of the titration curve where the pH changes rapidly, making it easy to detect the completion of the titration.

Equivalence Point

The equivalence point in a titration is reached when the amount of titrant added is stoichiometrically equivalent to the amount of substance originally present in the analyte. In the case of a weak acid-strong base titration, the equivalence point is characterized by the complete conversion of the weak acid to its conjugate base, and the pH at this point is determined by the hydrolysis of the conjugate base in water.

Buffer Region and Henderson-Hasselbalch Equation

During the titration, especially near the half-equivalence point, the solution behaves as a buffer containing appreciable amounts of both the weak acid and its conjugate base. The Henderson-Hasselbalch equation relates the pH of this buffer system to the pKa of the acid and the ratio of the concentrations of the conjugate base to the acid, providing a straightforward method to calculate pH in this region.

Weak Acid-Strong Base Titration

In a weak acid-strong base titration, the weak acid does not completely dissociate in water, and the addition of the strong base gradually converts the weak acid to its conjugate base. This type of titration is characterized by a relatively gradual change in pH before the equivalence point, a pronounced buffering region, and a equivalence point whose pH is influenced by the hydrolysis of the conjugate base.

Acid-Base Titration

Acid-base titration is a quantitative analytical method used to determine the concentration of an acid or base in a solution by gradually adding a titrant of known concentration. The reaction between the acid and the base results in a characteristic titration curve that illustrates the change in pH as titrant is added, which helps in identifying important features like the equivalence point and buffer regions.

Titration Curve

The titration curve is a graphical representation of pH versus volume of titrant added. It highlights key regions such as the initial pH, buffer region, half-equivalence point, the sharp pH change near the equivalence point, and the post-equivalence region where excess titrant dominates the pH calculation. This curve is essential for understanding the progression of the titration process.

Transcript

00:01 Okay, this is another one of those marathon titration problems that the author seems to like.

00:08 First, it asks us to calculate the ph before any sodium hydroxide is added.

00:14 So that means the ph of the 25 -milliter solution of 0 .05 molar hydrosyanic acid.

00:22 To do that, the hydronium ion concentration can be estimated by taking the square root of k .a., which you'll need to look up, 4 .0 times 10 to the...

00:32 The negative 10 multiplied by the concentration of hydrocyanic acid and we get 4 .47 10 to the negative 6.

00:40 The ph then is going to be the negative log of this value or 5 .35.

00:48 Now for part b it asks for the ph at the halfway point of the titration.

00:53 At the halfway point of the titration the ph is going to be equal to pca so it'll simply be the negative log of 4 .0 times 10 to negative 10.

01:03 Or 9 .40.

01:05 I'm going to skip and go to d and then i'll come back to c.

01:10 The equivalence point volume can be calculated by taking the volume of hydrocyanic acid converted to liters, multiplied by the concentration of hydrocyanic acid to get moles hydrosyanic acid.

01:24 We recognize that the reaction is one to one.

01:28 So one mole of hydrocyanic acid requires one mole of n -a -o -h.

01:34 Then we'll use the molarity of n -a -o -h to convert moles n -a -o -h into liters, and this is actually n -a -o -h, and then it's 0 .1 -667 liters of n -a -o -h that is required to reach the equivalence point.

02:00 So going back to c, if we're 95 % of the way there, then we're still in the buffer region.

02:10 Is going to be equal to pca.

02:13 The 9 .40, of course, is the negative log of the ka value.

02:18 Plus the log of down in the denominator, we have our concentration of acid that we're starting with, which concentration of acid we're starting with, multiplied by its volume and liters to get moles of the hydrosyanic acid, minus 95 % or 0 .95 of the volume required to reach the equivalence point multiplied by its concentration.

02:44 This then is the moles of n -a -o -h that is added which will correspond to the moles decrease in hydrosyanic acid.

02:53 It'll also correspond to the moles increase in the conjugate base.

02:59 So we'll take this exact same expression here and put it in the numerator in order to get the moles of the weak base formed, the cyanide.

03:07 And we get a ph of 10 .6.

03:10 Then for e, the hydroxide concentration at the equivalence point can be estimated by taking the square root of kb.

03:21 Kb is going to be ka, which is the 4 .0 times 10 to negative 10 divided into the kw...

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