Imagine that you are in chemistry lab and need to make 1.00 of a solution with a pH of 2.50.
You have in front of you
100 of 7.00×10−2 , HCL
100 of 5.00×10−2 , and NaOH
plenty of distilled water.
You start to add HCL to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH . Once you realize your error, you assess the situation. You have 81.0 ml of HCl and 90.0ml of NaOH left in their original containers.
Assuming the final solution will be diluted to 1.00L, how much more HCl should you add to achieve the desired pH?
pH=log(H)
2=log H
H=1E-2=.01M
HCl net must be .01moles.
You have .009*7E-2 moles or .009*.063 moles
So additional moles must be .01-above.
additionvolumeHCL=additionalmole/7E-2 in liters.
check my thinking.
your answer is pretty cracked out, if you gona help, show your steps and help
To determine how much more HCl should be added to achieve the desired pH, we can use the concept of calculating moles and molarity.
1. Calculate the moles of HCl and NaOH present in the original quantities:
- Moles of HCl = volume (in liters) x molarity of HCl
= 0.100 L x 0.0700 M = 0.007 mol
- Moles of NaOH = volume (in liters) x molarity of NaOH
= 0.100 L x 0.0500 M = 0.005 mol
2. Take into account the accidental addition of NaOH:
- Moles of HCl remaining = initial moles of HCl - moles of NaOH accidentally added
= 0.007 mol - 0.090 L x 0.0500 M = 0.007 mol - 0.0045 mol = 0.0025 mol
3. Determine the volume of the desired 1.00 M HCl solution needed to achieve the desired pH:
- pH is a measure of the concentration of hydronium ions (H3O+). In a solution with a pH of 2.50:
- [H3O+] = 10^(-pH)
= 10^(-2.50) = 0.00316 M
- We want to achieve a final concentration of 0.00316 M H3O+ ions.
- Moles of H3O+ ions needed = volume (in liters) x [H3O+]
= 1.00 L x 0.00316 M = 0.00316 mol
4. Calculate the additional moles of HCl required:
- Additional moles of HCl = Moles of H3O+ ions needed - Moles of HCl remaining
= 0.00316 mol - 0.0025 mol = 0.00066 mol
5. Convert the additional moles of HCl to volume:
- Additional volume of HCl = Additional moles of HCl / molarity of HCl
= 0.00066 mol / 0.0700 M = 0.00943 L = 9.43 mL
Therefore, you would need to add an additional 9.43 mL of HCl to the solution in order to achieve the desired pH of 2.50.