For a chemistry lab the experiment was to carry out an acid-base titration to determine the exact concentration of a sodium hydroxide solution.
Two trials were completed and the data collected was: mass of weighing paper, mass of weighing paper + KHP, mass of KHP, initial buret reading, final buret reading, and volume of NaOH.
The molecular weight of KHP is 204.44 grams/mol.
I need to find the number of moles of KHP, the number of moles of NaOH and the concentration of NaOH but i am not sure how to do this. Please help,thank you!!
so for the moles of KHP it would be 1/204.44 = 0.00489 g/mol?
The volume of NaOH for trial 1 was 14.50 mL and for trial 2 it was 9.90mL. so for molesof NaOH would it be 39.998 which is the molecular weight og NaOH divided by the volume for each trial?
Then for the concentration of NaOH would i use the number of moles that I got and divide that by the volumes of NaOH?
Thank you for your help! I am not very good with this chemistry stuff.
It would have been better to have typed in the numbers. As it is I don't know if you are right or not but it appears you are on the right track on parts and not on the right track for parts.
mass KHP = (mass KHP + mass paper)-mass paper.
volume NaOH = final buret reading - initial buret reading.
Here is the equation.
KHP + NaOH ==> NaKP + H2O
I have for the molar mass of KHP = 204.22.
moles KHP = mass in grams KHP/204.22
moles NaOH = the same since the equation shows 1 mol KHP = 1 mol NaOH
M NaOH = moles NaOH/L NaOH
You know moles NaOH from above and L NaOH (you show 0.00990L and 0.01450)
Note:I think I answered this before but some of what I said must have been garbled. At any rate, it is important that you understand the concept of titration. You want to determine the M NaOH. So you use an acid that you can weigh exactly (KHP), then titrate with an indicator to the exact neutralization point. When you are there you know that moles NaOH exactly equal moles KHP. Then knowing moles NaOH and volume of NaOH you can calculate M NaOH. The KHP is called a primary standard because you can weigh a sample of 100% KHP and know the moles that are there. You CANNOT weigh NaOH and know the number of moles because NaOH absorbs water and CO2 from the air; thus, NaOH cannot be used as a primary standard. I hope this helps.
yes thank you i understand the parts for moles of KHP and NaOH. So the mw of KHP was given to me which was 204.44g/mol so for moles of KHP i did 1/204.44 and got 0.004 and then for NaOH i got 0.004 as well. Then for the concentration of NaOH would i do 0.004/0.0145 and 0.004/0.00990 ?
Why are you throwing perfectly good numbers away?. I assume the 1 you are using comes from 1.00 grams KHP. Then 1.00/204.44 = 0.004891 moles which you should round to the appropriate number of significant figures. But you apparently just threw away the 891.
So (NaOH) = 0.004891/0.0145 and 0.004891/0.00990. I seem to remember you used 1.00 as grams which gives you three s.f. so the final answer should be rounded to 3 s.f.
Your precision is not very good if this is for the same NaOH solution. For whatever it's worth a VERY GOOD analytical chemist can duplicate results to about 2 parts per thousand for an acid/base titration. In real numbers, if the actual value is 0.1000M, a chemist might get 0.0999M to 0.1001M
To find the number of moles of KHP (potassium hydrogen phthalate), you will need to use the mass of KHP and its molecular weight.
First, let's calculate the number of moles of KHP. You correctly identified the formula for moles: mass/molecular weight.
In your case, you mentioned the mass of KHP. Let's assume trial 1 had a mass of KHP of 1.0 grams. The molecular weight you provided is correct at 204.44 grams/mol.
So, the moles of KHP in trial 1 would be 1.0 g / 204.44 g/mol = 0.00489 mol, which is correct.
Now, let's calculate the number of moles of NaOH.
Using the volume of NaOH, we can calculate the moles using the relationship: moles = volume * concentration.
In trial 1, if the volume of NaOH is 14.50 mL, we need to convert it to liters for the calculation. There are 1000 mL in 1 L, so 14.50 mL is 14.50/1000 = 0.0145 L.
Next, we divide the volume by the molecular weight of NaOH (39.998 g/mol) to get moles: 0.0145 L * (39.998 g/mol) = 0.5793 mol of NaOH for trial 1.
For trial 2, using a volume of 9.90 mL, we follow the same calculations: 9.90 mL / 1000 = 0.0099 L, then 0.0099 L * (39.998 g/mol) = 0.3960 mol of NaOH for trial 2.
Finally, to find the concentration of NaOH, we divide the moles of NaOH by the volume of NaOH.
For trial 1: 0.5793 mol / 0.0145 L = 39.93 mol/L or 39.93 M (Molarity represents moles of solute per liter of solution).
For trial 2: 0.3960 mol / 0.0099 L = 40.00 mol/L or 40.00 M (rounded to two decimal places).
So, the concentrations of NaOH for trials 1 and 2 are approximately 39.93 M and 40.00 M, respectively.
It's important to note that these calculations assume that the reaction between KHP and NaOH is stoichiometric (1:1 ratio).