Use the half-reaction method to balance the equation for the conversion of ethanol to acetic aci

Use the half-reaction method to balance the equation for the...

Key Concepts

Oxidation and Reduction Processes

Recognizing which species is oxidized and which is reduced is key to the half-reaction method. The oxidation process involves the loss of electrons (typically seen in the conversion of an organic molecule to a higher oxidation state compound), while the reduction process involves the gain of electrons (commonly observed in the reduction of a metal species in a dichromate ion).

Balancing in Acidic Solution

When balancing redox reactions in an acidic medium, additional steps are required to account for hydrogen ions (H+) and water molecules. By systematically adding H2O, H+, and electrons to balance oxygen and hydrogen atoms, the overall reaction is balanced while maintaining the proper charge balance.

Redox Reactions

Redox (reduction-oxidation) reactions involve the transfer of electrons between chemical species, where one substance is oxidized (loses electrons) and another is reduced (gains electrons). Understanding electron transfer is crucial to determining which species donate or accept electrons in a reaction.

Half-Reaction Method

The half-reaction method is a systematic approach to balancing redox reactions by separating the overall reaction into two individual half-reactions: one for oxidation and one for reduction. Each half-reaction is balanced independently for mass and charge, and then combined to obtain the overall balanced equation.

Transcript

00:01 Question 139 is a weak acid, strong base titration question, where, after adding certain volumes of the titrant sodium hydroxide to our analyte, hydrofluoric acid, we will calculate the ph.

00:19 This is our titration reaction.

00:21 It's one -to -one with hf and sodium hydroxide.

00:25 The first question asks, what is the ph of the hf solution before the titrant is asked? to carry out this calculation, we simply need to determine the hydronium concentration of the weak acid solution, which is equal to the square root of the concentration of hf, multiplied by the k .a.

00:50 Of h .f.

00:52 And that square root gives us our hydronium concentration.

00:56 We take the negative log of this hydrogenium concentration to get us our ph.

01:00 This is the equation that we use when we just have a weak acid in solution.

01:06 Hydronium concentration is equal to the square root of the weak acid concentration multiplied by its k -a value, and we get a ph of 1 .88.

01:16 Now for part b, it wants to know how many milliliters of titrant are required to reach the equivalence point.

01:22 To answer this part of the question, we will take the volume of the hf that we are starting with, 35 milliliters, converted to liters as 0 .035 liters.

01:34 We multiply it by the molarity, which is 0 .250 molar.

01:39 In order to convert the liters into moles of hf, we recognize the stoichiometry of the reaction was one to one, so one mole hf requires one mole naah.

01:52 Then we can go from moles nahoh to the leaders of nahoh solution that is serving as the titrant by dividing, the molarity of the n .a .o .h.

02:02 Solution so that the moles n .a .o .h cancel and we're left with liters n .a .o .h.

02:08 And then last of all, multiply it by a thousand to get the volume in milliliters, 57 .11 milliliters n .a .o .h.

02:18 Part c asks us to determine the ph at 0 .5 millilitres before the equivalence point.

02:25 Well, if this is the equivalence point volume, then we take off 0 .5 from that, and that will be the volume for which we need to calculate the ph.

02:35 Anytime we are post -initial and pre -equivalence, we have a buffer solution.

02:43 We still have some weak acid, and we have formed some weak base.

02:47 To calculate the ph of a buffer solution, we'll use the henderson -hasselbalch equation, where ph is equal to p -ka, which is the negative log of the k -a value for the weak acid, which in this case, is hf plus the log of the moles of base over the moles of acid or the concentration of base over the concentration of acid.

03:08 It's going to be easier to use moles, so that's what i'm going to use here.

03:15 So this is the volume of the equivalence point.

03:19 We subtract off 0 .5 milliliters, which will be 0 .005 liters, just like the 57 .111 milliliters is now 0 .05711 liters.

03:34 So this is the volume in liters that is 0 .5 milliliters pre, just 0 .5 milliliters before the equivalence point.

03:43 If we take the volume and we multiply it by the concentration of the sodium hydroxide that we have added, this will give us the moles of sodium hydroxide added.

03:52 And because the titration reaction tells us that for every mole of sodium hydroxide we add, we get the conjugate base f minus.

03:59 This is also the moles of f minus is formed.

04:03 We then go down to the denominator and we'll take the volume, 35 milliliters or 0 .035 liters, multiplied by the concentration of hf .25 molar.

04:17 This will give us the moles of hf that we start with, but as we carry out the titration, the hf is consumed and it's turned into the f minus.

04:26 So the amount of hf consumed will be equal to the moles of n .a .o .h.

04:33 That we added.

04:34 And the moles of n .a .o .h.

04:35 That we just calculated up here.

04:38 It was this difference in volume multiplied by the molarity.

04:41 Same difference in volume multiplied by the molarity...