A student titrates 25.00 mL of an HCl solution of unknown concentration with standardized 0.1050 M NaOH using phenolphthalein indicator. The endpoint (faint pink) is reached at 23.42 mL of base. A pH probe is used in a parallel run to record the titration curve, which shows a near-vertical jump centered at pH ~7.
What is the concentration of the HCl solution?
- A
0.112 M
- B
0.0246 M
- Ccheck_circle
0.0984 M
- D
0.1050 M
Explanation
Moles NaOH = 0.02342 * 0.1050 = 2.459e-3 mol = moles HCl. [HCl] = 2.459e-3 / 0.02500 = 0.0984 M.