Hi, I did the acid-base titration lab with HCl and 0.5M NaOH.
I need to calculate the moles of NaOH from molarity of NaOH and the average volume used.
My English isn't good but is it asking me to find the moles of NaOH using the molarity of NaOH and the average volume used correct?
This is my table.
Molarity of NaOH used: Trial 1
Initial volume of NaOH (mL): 27.10
Final volume of NaOH (mL): 38.90
Volume of NaOH used (mL): 11.80
Trial 2
10.80
22.60
11.80
Trial 3
22.60
36.70
14.10
My average volume of NaOH:
11.80+14.10/2=12.95mL
Then for some reason I am really confused even though I know it should be very simple... Please help me, thank you kindly.
The average volume is not right because you threw away one of the 11.8 volumes.
Volumes are 11.80, 11.80 abd 14.10
Sum those and divide by 3.
Then mols NaOH = M NaOH x L NaOH = ?
I thought we only add the first two because the first one was purposely overshot?
0.5M NaOH × 12.56666667= 6.283333333
=6.3 moles
Thank you for replying!
Yes, you are correct. The question is asking you to calculate the moles of NaOH using the molarity of NaOH and the average volume used.
To calculate the moles of NaOH, you can use the formula:
moles = molarity * volume (in liters)
First, convert the average volume used from milliliters to liters:
Average volume used = 12.95 mL = 12.95/1000 L = 0.01295 L
Next, use the formula to calculate the moles of NaOH:
moles = 0.5 M * 0.01295 L
Calculating this will give you the number of moles of NaOH used in the experiment.
No problem, I'm here to help! Your understanding is correct. The question is asking you to calculate the number of moles of NaOH using the molarity of NaOH and the average volume used.
To calculate the moles of NaOH, you can use the formula:
moles = molarity x volume (in liters)
First, let's convert the average volume of NaOH used from milliliters to liters. In this case, you have an average volume of 12.95 mL.
To convert 12.95 mL to liters, divide by 1000:
12.95 mL ÷ 1000 = 0.01295 L
Now, you can calculate the moles of NaOH using the molarity of NaOH and the converted volume:
moles = 0.5 M x 0.01295 L
moles = 0.006475 mol
Therefore, the moles of NaOH using the given molarity and average volume is approximately 0.006475 mol.
I hope this helps! Let me know if you have any further questions.