A buffer was prepared by mixing 0.1 M ch3cooh with 0.1M ch3cooNa in the ratio 8:2. Calculate ph after the addition of 2mL of 0.1 M HCl?
When you do the 8:2 ratio and add 2 mL of 0.1 M HCl, the buffer is gone and you are left with 1*10^-3 moles of ch3cooh. How would you calculate the pH after that?
the total volume should be 10 mL so the ratio 8mL:2mL before the addition of HCl
2 answers
You didn't read my earlier response? I guess it's just easier to repost and hope someone will work the problem for you.
You are correct about the buffer salt being consumed and that there are 0.001 mole of HOAc remaining. Now put this in 12 ml solution and calculate the [HOAc] for the resulting solution. [HOAc] = (0.003mol/0.012L) = 0.083M.
Given Ka = 1.8x10^-5 & [HOAc] = 0.083M
=> [H^+] = Sqr-Root(Ka x [HOAc])
pH = -log[H^+]. Good Luck
Given Ka = 1.8x10^-5 & [HOAc] = 0.083M
=> [H^+] = Sqr-Root(Ka x [HOAc])
pH = -log[H^+]. Good Luck
1 answer