A 10.0-mL sample of 0.250 MHNO3 solution is titrated with 0.100 M KOH solution. Calculate the pH

A 10.0-mL sample of 0.250 MHNO3 solution is titrated with...

Key Concepts

Acid-Base Titration

Acid-base titration involves the gradual addition of a titrant (acid or base) of known concentration to a solution containing the analyte in order to determine its concentration. It relies on the neutralization reaction between the acid and the base, with the reaction proceeding until a specific endpoint is reached, often identified by a change in pH or by an indicator.

Neutralization Reaction Stoichiometry

The neutralization reaction is based on the stoichiometric relationship between the acid and the base. For strong acids and strong bases, this reaction occurs in a 1:1 mole ratio, meaning that one mole of acid reacts completely with one mole of base. This concept is critical for determining how much titrant is required to reach the equivalence point and for calculating the concentration of the remaining species after partial neutralization.

Equivalence Point

The equivalence point in a titration is the stage at which the quantity of titrant added is just enough to completely neutralize the analyte present in the solution. At this point, the number of moles of acid equals the number of moles of base, making it a key reference marker for calculating pH changes before and after neutralization and understanding the progression of the titration.

Excess Reactant Analysis

When titrating an acid with a base (or vice versa), determining whether there is an excess of acid or base after a given volume of titrant is added is essential. This concept involves calculating the remaining moles of acid or base once the neutralization reaction has taken place and directly affects the calculation of the pH of the solution.

pH Calculation in Strong Acid-Base Systems

In titrations involving strong acids and strong bases, both substances dissociate completely in water. The pH of the solution is determined from the concentration of the excess hydrogen ions (in the case of excess acid) or hydroxide ions (in the case of excess base) present in the solution after the neutralization reaction. The relationship pH = -log[H?] is used to convert the concentration of hydrogen ions into pH, and similarly pOH = -log[OH?] followed by pH = 14 - pOH when hydroxide ions are in excess.

Transcript

00:01 In this problem, we are asked to consider a titration, and we're going to figure out the ph at various points in this titration.

00:09 The information we're given.

00:11 We have a 10 -millimeter sample of a 0 .250 molar nitric acid.

00:27 This will stay the same.

00:29 That is a constant.

00:33 And it's titrated with 0 .100 molar k -oh.

00:45 We're going to calculate the kh for five different amounts of base added.

00:51 And i'm going to do each one of these separately.

00:57 I actually did them in a table in my book and then went, or in my notebook, then went back and forth.

01:07 I think i'm going to do each one separately.

01:10 So let's do a.

01:12 For a, and i'm going to do each of these in the same way, i'm going to write the milliliters of hn -o -3, the milliliters of k -o -h, and the total volume.

01:45 So on this, we were given 10 .0.

01:48 That was right here.

01:50 It's the same for all of these.

01:51 And for our first trial, are given as 20 .0 milliliters.

01:55 This is they're given for trial a.

01:58 And total is 30 milliliters.

02:06 That's our total volume.

02:09 The moles of h plus will be the same for every trial.

02:15 In order to get that, we're going to take 0 .250 molar times 0 .010 liters.

02:30 So this value will be 0 .00250 moles.

02:40 Of hydrogen.

02:42 This will be the same for every trial.

02:45 This will be the same for every trial.

02:50 Next we're going to figure out how many moles of oh we have.

03:00 Do that right here.

03:05 And to figure our moles of oh, we're simply going to take our molarity, which was given as 0 .10 times the volume.

03:15 In this case, it was 0 .020.

03:22 And that gave us 0 .0020.

03:31 I need one more sig -figure.

03:33 On that.

03:45 That's moles.

03:46 Okay.

03:48 When we have that, we're going to start our calculation.

03:50 I'm going to change colors for the calculation.

03:53 For the calculation, we're going to take the larger of the two values, the larger of these two values.

04:05 So we're going to take 0 .00250 moles of h plus minus 0 .00200 moles of oh minus, which is a 0 .00 ,000, moles of oh minus, which is a lot of, which gives me 5 .0 times 10 to the minus 4th moles.

04:37 Then i'm going to divide that number, my 5 .0 times 10 to the minus 4th moles, divided by my total volume to get my molarity.

04:55 And that will be 1 .66 times 10 to the minus 2 molar in h plus.

05:05 Taking the negative log of 1 .66 times 10 to the minus 2 molar in h plus.

05:07 Taking the negative log of 1 .66 times times 10 to the minus 2, we get a ph of 1 .78.

05:23 And that is number one.

05:30 Now let's do our second calculation.

05:33 In our second calculation, again, we have our 10 milliliters of our hn03.

05:50 We have 24 .9 milliliters this time of our k -oh, which gives us a total volume of 34 .9 milliliter.

06:11 There's my volumes.

06:13 If you recall the concentration or the number of moles of hydrogen is the same for every trial, and that is 0 .00250.

06:35 For this, our moles for, boy, i keep pulling that down there, our moles of oh minus will be equal to zero.

06:45 I'm going to leave my units off.

06:47 There's my molarity times my volume.

06:56 I'm not going to be able to get that on there.

06:58 I have to do this down here.

07:12 0 .100 times 0 .0249 equals 0 .00249 moles.

07:33 Again, our next calculation, we will take our larger of these two numbers and then subtract the smaller.

07:43 Larger, minus smaller.

08:10 And from that i get, i should have probably put these in scientific notation, but who cares? and that's moles of h plus that are in excess.

08:26 To get my malaria, we're simply going to divide that, divided by my total moles, which was 0 .0349...

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