For the titration of 25.00 mL of 0.150 M HCl with 0.250 M NaoH, calculate (a) the initial pH; (b) the pH when neutralization is 50% complete; (c) the pH when neutralization is 100% complete; and (d) the pH when 1.00 mL of NaOH is added beyond the equivalence point.
4 answers
Here is what you need to do. Determine what you have in solution at each of the four points. To do that, determine moles HCl initially, moles NaOH added, subtract to see which is in excess and go from there. Post your work if you get stuck.
I determined the number of millimoles of HCl initially to be 3.75.
How do I determine how much moles of NaOH are added at 50% completion for example because the volume isn't given.
How do I determine how much moles of NaOH are added at 50% completion for example because the volume isn't given.
The reaction is:
HCl + NaOH --> NaCl + H2O
For 100% neutralization, moles NaOH needed is 3.75.
For 50% neutralization, you use half of 3.75 millimoles NaOH. At that point, 50% of the HCl is still there.
[H+] = conc. of HCl = (moles HCl)/(Total volume)*
*Note: Total volume (L) = 0.02500L+(0.001875mol/(0.250mol/L)
HCl + NaOH --> NaCl + H2O
For 100% neutralization, moles NaOH needed is 3.75.
For 50% neutralization, you use half of 3.75 millimoles NaOH. At that point, 50% of the HCl is still there.
[H+] = conc. of HCl = (moles HCl)/(Total volume)*
*Note: Total volume (L) = 0.02500L+(0.001875mol/(0.250mol/L)
Ok Thanks a lot guys.
1 answer