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A student titrates an unknown monoprotic acid with a NaOH solution from a buret. After the addition of $12.35 \mathrm{mL}$ of $\mathrm{NaOH},$ the $\mathrm{pH}$ of the solution read 5.22. The equivalence point is reached at $24.70 \mathrm{mL}$ of NaOH. What is the $K_{\mathrm{a}}$ of the acid?

$$K_{\mathrm{r}}=6.026 \cdot 10^{-6}$$

Chemistry 102

Chapter 16

Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria

Aqueous Equilibria

Rice University

Drexel University

University of Toronto

Lectures

00:41

In chemistry, an ion is an atom or molecule that has a non-zero net electric charge. The name was coined by John Dalton for ions in 1808, and later expanded to include molecules in 1834.

24:14

In chemistry, a buffer is a solution that resists changes in pH. Buffers are used to maintain a stable pH in a solution. Buffers are solutions of a weak acid and its conjugate base or a weak base and its conjugate acid, usually in the form of a salt of the conjugate base or acid. Buffers have the property that a small change in the amount of strong acid or strong base added to them results in a much larger change in pH. The resistance of a buffer solution to pH change is due to the fact that the process of adding acid or base to the solution is slow compared to the rate at which the pH changes. In addition to this buffering action, the inclusion of the conjugate base or acid also slows the process of pH change by the mechanism of the Henderson–Hasselbalch equation. Buffers are most commonly found in aqueous solutions.

01:11

A student titrates an unkn…

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01:45

A student pipets 10.00 mL …

02:49

When $40.00 \mathrm{~mL}$ …

03:27

It is found that it takes …

05:10

The titration of 0.02500 L…

04:05

A sample of a certain mono…

12:53

05:30

A weak monoprotic acid is …

00:52

01:44

A $0.288-\mathrm{g}$ sampl…

01:58

the solution for this is dead. This is it. Penetration off a weak acid, that is monopoly. Take answered with a stronger base that is, sodium hydroxide for this type of titillation, the pH at the half way point toe. The equivalent point is equal toe peak off the asset. That means the value off PK off the A normal Mona Protic acid is 5.52 Therefore, the explanation for the north this okay, is, uh, try tradition offer week answered. That is Mona Protein A sued with a strong based, which is sodium hydroxide. Okay, for these type off dietitians, the pH at the half a point. Yeah, door Mhm. The equivalence point is equal toe BK off the Massoud. That means the value PK yeah, off the unknown Monta project acid. Yeah, yes. 5.2 rich correspond toe. Well, you or okay is given as that PK equal to minus law. Okay, 5.2 equal to minus law. Okay, so the K they will be equal to 10 to the power minus 5.2 by solving this and it will be equal to 6.0 three multiplied by 10 to re power minus six. Therefore, the value off a four unknown Monath protic acid is x 40.0 three. Multiply by 10 to repair minus six. So this is the solution off. The answer that this is a titillation off a week, as in that is Mona Pro two captured with a strong they sodium hydroxide for these type of titles. And the pH at the half way point to the equivalence point is equal toe PK off the asset. That means the value peak off the unknown. Mono protic acid is 5.22 which corresponds to the value. Okay, is giving us a PK is going tomorrow. Okay. And by solving this here, we get the value. Okay, It is time to depart minus 5.21 which is equal to zero point. So D, which is equal to 6.23 10 20. Power minus 66.3 to 24 menacing. Therefore, the value off K for unknown mono protic as it is 6.3 multiplied by 10 to the power six. Thank you

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