Can someone help me with this question please?
What is the pH at each of the points in the titration of 25.00 mL of 0.1000 MHCl by 0.1000 M NaOH:i) After adding 25.00 mL NaOH ii) After adding 26.00 mL NaOH
CHEM 102 - DrBob222, Friday, August 19, 2011 at 8:35pm
This is a titration problem of a strong acid with a strong base.
HCl + NaOH ==> NaCl + H2O
These titrations can be divided into four distinct parts.
a. At the beginning. pH is determined by the pure HCl with no NaOH added.
b. At the equivalence point. pH is determined by the hydrolysis of the salt. IN this case, the salt is NaCl which is neutral so the pH is 7.
c. All points between a and b. moles HCl - moles NaOH = moles HCl remaining. pH = -log(H^+).
d. All points after b. All you have in the solution is excess NaOH. Total moles NaOH added - moles HCl initially = moles excess NaOH and M NaOH = moles NaOH/total volume. Then pOH = -log(OH^-) and pH + pOH = pKw 14.
Post your work if you get stuck.