# Concentration of HCl

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I was practicing a titration lab today that I have to do next week. I was titrating HCl with NaOH from a buret. Using an indicator to obseve the endpoint, I found that 18.32ml of NaOH was needed to react with 25ml of HCl. The concentration of the NaOH was .08045 M, and I needed to find the concentration of the HCl. Using the equation $M_1V_1 = M_2V_2$ and converting the ml to liters, I found the final concentration of NaOH was .1277 M, and so I assumed that the HCl would be of the same concentration, but I'm supposed to get .05927 M for HCl. For M1 I have .08045 M NaOH, V1 is .0495L, what I started with, and V2 is .03118L, what I had left. I'm not sure if I'm going about the calculations wrong, or if I maybe screwed up the experiment itself. Any suggestions are appreciated.

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It's about stoichiometry, but it's also about X (sub:needed) and Y (sub:needed)

So much of X is needed to neutralize Y. When neutralization occurs, color change occurs. Sometimes screwups happen if you don't correctly titrate. A way this is possible is by not dropping the liquid. Some people squirt the liquid too quickly. A skilled person tends to drop, squirt, drop, slower squirt, and when it begins to just slightly change, drop. It helps to do an experiment like this more than once for comparison.

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C1V1 = C2V2 - yes, but 1 means NaOH solution and 2 means HCl solution. The equation looks the same as in teh case of dilution, but it is not the same. So what you have is 18.32ml*0.08045M=25ml*C(HCl). Solve for C(HCl). Looks to me like your error was about 0.55%.

Borek

--

Stoichiometry calculator

www.pH-meter.info/pH-electrode

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AHHH, I see what I did wrong. I really appreciate the replies, thank you

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It's one of those decent experiments a person can use hands-on mathematics from empirical observations. Just keep working on applying theory to reality, and you will get the hang of it.

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Until you acquire enough experience with the various indicators its best to titrate at a rate of a drop or two per second until you see you are nearing the endpoint. When you are nearing the endpoint slow down the rate at which you are adding titrant. You can add a drop, swirl the Erlenmeyer, add another drop, swirl, etc. You can even add a half drop and touch the inside of the Erlenmeyer to the hanging drop. Just make sure to tip the Erlenmeyer to collect the titrant that’s holding onto the sides of the Erlenmeyer. A rule of thumb to remember during analytical work is you’re chasing molecules.

If you eventually pursue chemistry as a profession it’s important to realize that your company is paying you per hour. They want to get as much work out of you in that hour as possible. So you can’t titrate at a rate of a drop per second. What I’ve learned to do is open the stopcock to allow a slow stream of titrant into the flask. I swirl the flask once every couple seconds. When you are near the endpoint you’ll see the color starts to hang around longer than usual. That’s when I slow down and add drop by drop.

Lastly, burets are most accurate in the middle range of the graduations. So a 50mL buret is most accurate around the 20-40mL mark.

As for your calculations it appears to be common practice for chemistry students to memorize formulas. I’d rather see teachers teach students a tool called dimensional analysis.

Here is how I’d workout the problem. Now I found your original paragraph above a bit confusing so I hope I understood correctly.

18.32mL x ( 0.08045moles NaOH / L ) x ( 1mole HCl / 1 mole NaOH ) x ( 1/25mL HCl ) = 0.059 M HCl.

Below is a decent link.

http://www.chem.tamu.edu/class/fyp/mathrev/mr-da.html

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I appreciate the responce, I understand it now. I was tested on it yesterday, and I was able to do 2 titrations in under an hour haha, I dont know if thats good, but it's good for me.

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