okay, so i have NO idea how to figure out the ratio of hydrogen ions to hydroxide ions .. and i need a specific formula and everything
if anyone answers back i will give you the information
In what?
chemistry titration ? does that help
oh and grade 11
Perhaps you are Redfire. I've not seen another chemistry question from grade 11. Give us the info and we can help but we need the info. We aren't clairvoyant.
yeah i am .. embarassing, but i really need help if you go look at my other one there is all the information
or here ...
Follinwing this procedure:
1. Obtain a burette. Rinse it by pouring water through it. Add 15 ml of the HCl into the burette. Hold the burette horizontal and swirl it to further prepare it for the experiment. Pour this acid down the drain.
2. Place the burette in a clamp. With a funnel slowly add approx. 50 mL of the HCL to the burette. Make sure that it flows freely by discarding about 2 mL of acid by opening the valve.
3. Use a pipette to transfer 10 mL of your dilute base to a clean flask (i'm assuming the bas is NaOH).
4. We are using the acid in the burette because the base clogs the valves. You will use a fresh 10 ml of base in the flask for each trial. You will add the indicator provided, add just enough drops to make the colour easy to distiguish your base. You will do at least 3 trials. The first one is a calibration trial. IT is done fairly fast to get a rough idea of how much acid is neede to neutralize the 10 ml of base. The following trials are done more carefully. As your titration approaches neutralization, slow the valve down to a drop by drop speed.
Do enough trials so you have a percent difference of 5%.
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So i did all of this in class, and i did 3 trials obviously, and i did not got a 5% reading, but that was because my indicator was not very well and my teacher said that was okay as long as i did everything else right.
My Initial reading for the calibration run was 1.2 ml, for trial 1 it was 2.0 ml and for trial 2 it was 1.2 ml. My final reading for the calibration run was 37.2 ml, for trial 1 it was 39.1 ml and for trial 2 it was 31.8 ml. The final readings for calibration run were 36.0 ml and for trial 1 it was 37.1 and for trial 2 it was 30.6
So what i need to know:
1. what is the ratio of hydrogen ions to hydroxide ions?
Ratio?
(H^+)/(OH^-) = (mL acid/mL base)
Thats it for THIS PARTICULAR ACID AND PARTICULAR BASE. You don't provide any numbers for molarity; therefore, you can't do mole ratios.
well theres another part but i didn't know if it had anything to do with the questions ..
does this help?
concentration of stock NaOH = 1.000M
Volume of Stock NaOH needed = 10ml
volume of water needed = 190 ml
volume of dilute NaOH made 200 ml
concentration of dilute NaOH 0.05 M
M(HCl)*L(HCl) = M(NaOH)*L(NaOH)
M(HCl)*L(HCl) = mols HCl.
M(NaOH)*L(NaOH) = mols NaOH
You can obtain the molar ratio by dividing one by the other. If you do the molar ratio, what number should you obtain if it is obtained by titration. Should it be 1.00?