Question

a strong acid-strong base titration was performed as follows: 10.00 ml of 0.10m hcl were collected in an erlenmeyer flask, and then diluted to 100 ml with distilled water. after that, two drops of indicator were added. then, the buret was filled with 0.100 m naoh. the titration was started by slowly adding naoh to hcl. (note: the volume of two drops indicator is ignored as it is so tiny compared with 100 ml.) what is the ph of the solution in the erlenmeyer flask after adding 2.00 ml of 0.100m naoh?

Answer 1

After adding 2.00 mL of 0.100 M NaOH to the diluted solution of HCl, the remaining moles of HCl are 0.0008 moles, giving a concentration of 0.00784 M in a total solution volume of 102 mL. The pH of this solution can be calculated as approximately pH = 2.11.

Step-by-step explanation:

The student performed a titration of a strong acid (HCl) with a strong base (NaOH). Initially, 10.00 mL of 0.10 M HCl were diluted to 100 mL, and the titration was begun with 0.100 M NaOH. The question is to calculate the pH after adding 2.00 mL of NaOH. To calculate the pH at this point, we need to consider the amount of moles of HCl initially present and the amount of moles of NaOH added.

Initially, there were 10.00 mL of 0.10 M HCl, which equates to 0.0010 moles of HCl (10.00 mL × 0.10 mol/L). After adding 2.00 mL of 0.100 M NaOH (2.00 mL × 0.100 mol/L), we have added 0.0002 moles of NaOH, which neutralizes some of the HCl. The remaining moles of HCl are 0.0008 moles (0.0010 moles - 0.0002 moles).

Since the total volume of the solution is 100 mL + 2 mL = 102 mL or 0.102 L, the concentration of HCl after adding the NaOH is 0.0008 moles / 0.102 L = 0.00784 M. The pH of this acidic solution can be calculated using the formula pH = -log[H+], where [H+] is the molar concentration of H+ ions, which in this case is equal to the concentration of HCl since it is a strong acid and dissociates completely.

Thus, the pH can be calculated as pH = -log(0.00784), which is approximately pH = 2.11.