Imagine that you are in chemistry lab and need to make 1.00 of a solution with a pH of 2.50.

Question

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?

Answer 1

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.

Answer 2

your answer is pretty cracked out, if you gona help, show your steps and help