Calculate the pH of the following strong acid solutions:5.00mL of 1.00 M HCl diluted to .750L?
Thanks so much people!!! I had it all the way up to .005 part!! Gosh I love yahoO!!
- skipperLv 71 decade agoFavorite Answer
5.00 mL x 1.00 M = 5 millimoles of HCl ( 0.005 moles)
0.005 moles HCl / 0.750 L = 0.00667 M
pH = -log(0.00667) = 2.18
- Anonymous1 decade ago
1.00 M = 1 mole per liter
5 mL = 0.005 L so there is 0.005 moles of the stuff
now you have 0.005 mole in .750 liters so you divide:
0.005/.750 and you get .001/.15 or (1/150) M thats the new concentration
now because this is a strong acid the hydrogen completely dissociates and you can calculate the pH with this equation:
pH = -log(H+) = -log(1/150) = 2.18 thats your answer please dont drink this solution you already have enough HCl in your stomach!
- jonLv 41 decade ago
First find the concentration of HCl after diluting the original solution:
No of moles of HCl = 1.00 * 5 / 1000 = 0.005
0.005 moles in 0.75L = 0.005 * 1 / 0.75 = 0.00667 moles/litre
So, [HCl] = 0.00667 M
Since HCl is a strong acid (completely dissociated into ions),
HCl ---> H^+ + Cl^-
Then [H^+] = 0.00667 M
pH = -log[H^+] = - log(0.00667) = 2.18