posted by Cathy .
Can you help me with the following: Calculate the pH that results when 25 mL of .1M HCL is titrated with .1M NaOH solution run in from a buret at each of the stages of volume of .1 M NaOH add in 5.0 mL incrememnts beginning with 0 mL of NaOH
HCl + NaOH ==> NaCl + H2O
At zero mL. YOu have 0.1 M HCl so the pH is pH = -log(H^+). Since HCl is a strong acid, H^+ = 0.1 M and pH = 1.
All of the others are done this way:
a. moles HCl initially = M x L = 0.025 x 0.1 = 0.0025 moles HCl.
b. moles NaOH added. (at 5 mL this is 0.005L x 0.1 M = 0.0005 moles NaOH.)
c. Moles HCl are neutralized by moles NaOH. Start with 0.0025 HCl - 0.0005 NaOH = 0.0020 moles HCl left. (H^+) = 0.0020moles/L (25 mL from HCl + 5 mL of added NaOH = 30 mL = 0.030 L)
d.pH = -log(H^+). Substitute remaining H^+ and solve.
All of the points work the same way until reaching the equivalence point which is at 25 mL NaOH. Here the pH is determine by the hydrolysis of the salt. Neither Na^+ nor Cl^- are hydrolyzed; therefore, the pH is 7 from pure water.
Points after the equivalence point are just excess NaOH. Therefore, calculate excess OH and pOH from that and convert to pH.
what is the pH of a solution obtained by mixing 70 mL of 0.125M NaOH with 30ML of 0.105M of HCL