ACIDSBASE REACTIONS AND TITRATIONS 
One of chemistry's best tricks is to determine the concentration of various substances. Imagine buying vinegar and sometimes the acetic acid concentration is so weak that it tastes like water, and other times it is so strong that it burns the skin off the tongue and throat. So finding concentration is an important skill. Many substances that we use are either acidic or basic. Because acids and bases can neutralize each other, that gives us a strategy for finding the unknown concentration of an acid or a base when we know the concentration of the other one. 

The concentration of all of these inorganic acids could be determined by using a base of known concentration. One base commonly used is sodium hydroxide (NaOH).  
The concentration of all of these organic acids could be determined by using a base of known concentration. Again, the base commonly used is sodium hydroxide (NaOH) because it is a strong base and can pull off any of the acid hydrogens (shown in red) from these organic acids.  
The concentration of all of alkaline solutions or compounds can be determined by using an acid of known concentration. Hydrochloric acid (HCl) is often used to neutralize the OH ions (hydroxide ions) in alkaline solutions. Also the amount or concentration of nitrogen compounds that absorb H+ ions can also be measured using a strong acid like HCl.  

For example, in the last quiz, you did a problem about how lye (NaOH) and aspirin will neutralize each other. In other words, the OH from NaOH will combine with H+ coming off of aspirin (acetylsalicylic acid) to make H2O. Let's say we suspect the aspirin tablets do not contain 325mg of aspirin as they claim. If we trust that the NaOH is 100% lye as the lye bottle claims, we can use that to test the amount of aspirin in the tablets. One approach is to figure out equal quantities that will neutralize each other. We take 1 tablet of aspirin and see if the truly has 325mg. To know how many grams of NaOH it would take to neutralize this we have to count the aspirin molecules. So we use the molar mass of aspirin, which is 180.157 g/mole, to count the aspirin molecules. The count for NaOH needs to be the same if they are to neutralize each other. So we turn 325 milligrams of Aspirin into moles. That number of moles is the same for NaOH. Those moles get changed to grams of NaOH. 



So what we could do is dissolve both of these into some water, mix them, and then check the pH. If it is neutral, then we know that we had equal number (equal moles) of both NaOH and Aspirin. If the solution is alkaline, we know that there was more NaOH than Aspirin, meaning less than 325 mg of Aspirin was present. If the solution is acidic, then we know there was more than 325 mg of Aspirin present. In other words, there was more than 0.00180 moles of Aspirin because it neutralized the 0.00180 moles of NaOH and still had some Aspirin left over to make the solution acidic.  
Instead of using pH paper, we could add a pH indicator such as phenolphthalein. If the color of the solution is purplish, then there was less Aspirin than NaOH meaning there was less than the claimed 325 milligrams. If the solution is clear, then there is equal or more Aspirin than NaOH. The problem with both of these approaches is they don't tell you how much less or more is the Aspirin compared to NaOH. So we need a better approach. 

An improvement was to not to just dissolve the NaOH in water and dump it into a solution of the dissolved Aspirin all at once. The idea was to add it to the solution of dissolved Aspirin a little at a time. We call this method "tritration". "Titrate" comes from "Titer/Titre" which comes from the same word as "title". A title can signify a certain quality like the title of "Premium, champion, firstclass, etc". "titre" was a method used to report the quality of gold or silver. So the idea of determining the quality of gold or silver is similar to what we do now we use now to determine the quality (strength) of substance when doing titrations. Above we calculated that 0.0720 grams of NaOH will neutralize 1 Aspirin tablet. Since 0.0720 g is too small to weigh out, we can make up What we could do is take twice that weight of NaOH (2x0.0720g), which is 0.144 grams and dissolve it in 50mL of water. We could then add this solution of NaOH to the solution of 10 dissolved Aspirin tablets. After adding 25mL, there will be 0.0720 grams of NaOH added, and all of the Aspirin should be neutralized. If it gets neutralized before 25mL, then that means there is less Aspirin than the 325mg expected. If it takes more than 25mL, then the Aspirin present is more than 325mg. The amount of mL more or less than 25mL would let us figure out the actual amount of Aspirin present. 

The approach to calculate the amount of Aspirin present is to realize that equal moles of both Apsirin and NaOH will neutralize each other. So what you need is a known amount of NaOH in your solution. Above we said we added 0.144 grams. 0.144g x 1 mole per 40g NaOH gives us 0.00360 moles (Columns AD). Since we dissolved it in 50mL, that means we have 0.0360 moles per 50mL. Dividing the 50 into 0.0360, we get 0.000720 moles per mL (answer at H). That means for every mL we add we are adding 0.000720 moles of NaOH. Using a buret like shown above, we can easily measure a tenth of a mL. Let's say we added 23.5 mL of this solution to the solution with the dissolved Aspirins and that's when all of the Aspirin was neutralized. So for every mL added, we dripped in 0.000720 moles of NaOH which neutralized 0.000720 moles Aspirin. To figure that we do 23.5 mL x 0.000720moles per mL. Our answer is 0.01692 moles of NaOH added and the same number of moles of Aspirin neutralized. We confirmed that 0.01692 moles of Aspirin was present. We can convert that to grams using its molar mass (180.157 grams per mole) and we get the final grams of 0.305 grams (305mg) of Aspirin present in the solution of 1 dissovled tablet. So it is short by 20mg.


Let's say you reported to the manufacturer of the Aspirin tables that they were short on the amount of Aspirin that was claimed. They asked about your method and you said you used NaOH to test it. They then asked how did you know the concentration of the NaOH solution. You told them you weighed out 0.144 grams of NaOH and added water until you had 50.0 mL of solution. They said weighing NaOH is not accurate because NaOH exposed to the air will quickly absorb moisture in the air. So the weight of 0.144 grams is not just from NaOH but also water. So the actual mass of NaOH is less than 0.144grams. You then remember that when you left some of the NaOH granules on the table that they started to look wet. A few hours later, there was so much water that most of the granules were dissolved. So you realized they were correct. 

So the problem is that the solution of 0.0000720 moles per mL (0.0720 moles per liter) of NaOH is only approximate. Is there a way to make up a solution of NaOH and then measure its concentration accurately? Yes, there is. It's by using a known amount of acid. One acid that is commonly used is called potassium hydrogen phthalate or KHP for short. This organic acid is a solid and easy to weigh out. It also doesn't absorb water so when exposed to the air, its weight doesn't change. So the strategy is to first use the NaOH solution to neutralize a known amount of KHP. That will tell you just how much NaOH is in the solution. Then the NaOH solution with an accurately known concentration can be used to neutralize the Aspirin (acetylsalicylic acid) so we can get an accurate concentration of Aspirin. 

KHC_{8}H_{4}O_{4} > H^{+} + KC_{8}H_{4}O_{4}^{} 
To the left is the chemical equation for KHP releasing one hydrogen ion. That means each KHP molecule will neutralize one NaOH molecule because H+ + OH > HOH. 

Our task was to neutralize 1 Aspirin tablet that contain a total of 325mg of Aspirin. We calculated that was the same as 0.00180 moles of Aspirin. Instead of neutralizing Aspirin, we will neutralize KHP. So let's weigh out 0.001800 moles of KHP. We have to find molar mass into order to know how many grams that is. From the formula we can add up the atomic masses of all elements and find out that it weighs 204.22 g/mole. So we multiply 204.22 g/mole times the 0.001800 moles and the moles cancel giving us 0.3676 grams. So we weigh that out on an analytical balance and dissolve it in about 50 mL of water. The amount of water doesn't matter here. It's just to get the KHP to dissolve. The important thing is we know the exact amount of KHP in that water.  
Image by J.A.Freyre 
The procedure is to dissolve 0.3676 grams of KHP in about 50 mL of water and add a couple of drops of phenolphalein to tell us when all of the KHP is neutralized. The solution will turn purplish as soon as there's just a quarter drop NaOH more than what is needed to neutralize the KHP. The KHP solution is put into an Erlenmyer flask like show.


Previously we only made up 50mL of NaOH solution, but this time we want to have extra NaOH so we have enough to do tritrations several times. So let's make a liter of NaOH solution. Our target concentration before was 0.0720 moles per liter, but let's just round that up to 0.1 mole per liter. It's weight then is 0.1 moles x 40.0 grams per mole, which comes out to be 4 grams of NaOH. We weigh out about 4 grams knowing that the weight isn't accurate because NaOH absorbs water. We then dissolve that in 1 liter of water giving us approximately 0.1 moles per liter and put it in a container that has a cap or stopper. CO2 will neutralize the NaOH if left exposed to long. We pour some of this into a 50mL buret and record where the starting volume level. Remember burets have zero at the top. Let's say the level of the NaOH solution reads at the 2.42mL mark (estimated just past the 2.4 line).


We then start adding the NaOH slowly to the flask with the KHP. It's basically let it flow a drop at a time. When it gets close to the point where all of the KHP is neutralized the clear solution of KHP will start to turn purplish but then the color goes away. Each drop added will cause the purplish color to persist longer. At one point the purplish color does not go away. That's when you stop and read the level on the buret. The level has dropped to 19.96 mL mark. The difference from 2.42mL and 19.96mL is 17.54mL. That's the volume of NaOH that has neutralized all of the KHP. Now you can calculate how strong your NaOH solution really is. We know it is in the 0.l mole per liter vicinity, but this pins it down more exactly. Below are the calculations. 

Here are the calculations to determine an accurate concentration of the NaOH solution.


Now we can return to out task of determining if these tablets truly have 325mg of Aspirin in them. We do the procedure just like with did with KHP, but this time our unknown is the Aspirin and our known is the NaOH concentration. See below for calculations 

Here are the calculations to determine an accurate concentration of the NaOH solution. When working with concentrations, you can always multiply the volume times the concentration to get the amount (moles) in that volume. Notice how liters of NaOH cancel out as well as moles of NaOH and moles of Aspirin. The only unit left is grams of Aspirin. 0.3124 grams is 312.4 mg. That's 12.6 mg less than it should be. So the manufacturer really did short people on the amount. Not by much, but about 4% less. Of course their profit is up by 4%. When there's millions of dollars at stake, 4% is quite a bit.


Let's say this vinegar needs to be tested to see if it really is 5% acidity, which means it 5% w/v acetic acid. As you should know, 5% w/v means it has 5grams per 100mL. So lets say we measure 10.00mL of the vinegar and put it in a flask with a a few drops of phenolphthalein. Remember, phenolphthalein is clear in acidic solutions, but will turn purplish if the solution turns alkaline (basic). So we fill up the buret again with our 0.1026 M NaOH solution and start adding it to the vinegar solution drop by drop. You started at the 1.25mL mark but kept adding the NaOH solution all the weigh down past the 50mL mark and the solution was still clear. So you realized that it will take more NaOH solution than what our buret can measure before it neutralizes all of the acetic acid in the 10mL sample. So you realized the 10mL sample was too big. So the next time you only measure 3.00mL (0.003L) of vinegar. So you refill the buret with NaOH and the start level is 0.52mL. You add the NaOH to the vinegar drop by drop and it turns and stays purple when you reached the 27.74mL mark. So that means you added 27.22mL (27.740.52) of 0.1026M NaOH. That equals 0.02722 Liters. See below for calculations. Molar mass of acetic acid is 60.05  
The below are the calculations. It starts the same as the test for Aspirin, but we end with dividing by 3mL and multiplying by 100/100 because we are looking for concentration in grams per 100mL.
So this titration showed that the vinegar bottle had a little more than 5% acidity. It had 5.59%. 

Quiz on titration 

Let's say you tested a different brand and the titration took 25.75mL of 0.1026M NaOH. You also took a 3.00mL sample of the vinegar to test.
Problem 1: What is the grams of acetic acid per 100 mL of vinegar (M2)? Problem 3: What formula do you put in M2? Remember you don't include K3 (100) because that goes directly into M3. You also don't need to include any cells that have a "1" in them. Problem 4: If A2 goes higher, does that make the final grams (M2) higher or lower?


Oxalic acid is a toxic acid. A person drank some water that had been contaminated with radiator cleaner that contained oxalic acid. The doctors need to know what the concentration was of the oxalic acid in %w/v. You do a titration and came up with the following data: Notice you need the molar mass of oxalic acid and notice that 10.00mL of the oxalic acid solution was used in the titration. Also notice that oxalic acid releases 2 H+ ions.
Problem 5: How many grams of oxalic acid is there per mole (G2)? Problem 6: How many grams of oxalic acid per 100mL of the sample (M2)? Problem 7: What is the % w/v of oxalic acid in the sample? 
Here's a bottle of Milk of Magnesia (solution of magnesium hydroxide Mg(OH)_{2}. The label says 8% w/v. Again, that's just the label. Does it really contain that amount? We can use a known concentration of an acid to neutralize the magnesium hydroxide using titration methods. On hand we have a solution of 0.1124 molar (moles per liter) HCl. Like what was done with NaOH, we put the known concentraton of HCl solution into a 50mL buret. To know how big of a sample of the Milk of Magnesia to use for the titration, we need to do some preliminary calculations. Titrations are more accurate if you use about 20 or more mL of the titrant (solution in the buret). So let's figure the moles of H+ in 25 mL of the 0.1124 M HCl solution and how many mL of Milk of Magnesia would equal that. Since Mg(OH)_{2} has two OH per molecule it only takes half as many moles to neutralize the H+ from HCl.
Problem 8: How many moles of HCl is in 25mL of 0.1124 M HCl solution (H2)? Problem 9: How many mL of this 8% w/v Mg(OH)_{2} is needed to neutralized those 25mL of HCl (H5)? 

The above calculation shows that we need a little more than 1 mL of the 8% w/v Mg to use about 25mL of the HCl solution. So we will measure out as accurately as possible 1mL of the 8% w/v solution. Our volumetric pipette is accurate to the hundredth of a mL. So we measure out 1.00mL and put it in flask. We would add about 50mL of water so we can stir it easier and see the color change easier. We could use phenolphthalein as an indicator but it would make this milk of magnesia start out purplish and then go clear at the point of neutralization. Another indicator is called Methyl Orange. It will be start off orange in the alkaline pH of Mg(OH)_{2}, but then turn red when it gets neutralized and the HCl starts in excess. So the titration was done and the start level was 1.88mL and the end point was at 27.55mL. So the mL added was 25.67mL (22.551.88).  
Problem 10: (was 8) How many grams of magnesium hydroxide per mole of magnesium hydroxide (G2)? Problem 11: (was 9) How many grams of magnesium hydroxide per 100mL of the sample (M2 with 3 sig figs)? Problem 12: (was 10) What is the % w/v of magnesium hydroxide in the sample (3 sig figs)? 

Send your answers to chm151@chemistryland.com with subject line of "Titration quiz". 