A 50.0 mL sample of 0.50 M HCl is titrated with 0.50 M NaOH. After 28.0 mL of Na

Question

A 50.0 mL sample of 0.50 M HCl is titrated with 0.50 M NaOH. After 28.0 mL of NaOH have been added to the acid, which of the following statements is true? A. The pH of the solution is less than 7 B. The pH of the solution is equal to 7 C. The pH of the solution is greater than 7 D. Need more information A 50.0 mL sample of 0.50 M HCl is titrated with 0.50 M NaOH. After 28.0 mL of NaOH have been added to the acid, which of the following statements is true? A. The pH of the solution is less than 7 B. The pH of the solution is equal to 7 C. The pH of the solution is greater than 7 D. Need more information A. The pH of the solution is less than 7 B. The pH of the solution is equal to 7 C. The pH of the solution is greater than 7 D. Need more information

Explanation / Answer

mol of HCl = 50.0mL x 0.50M = 25.0 mmol
mol of NaOH = 28.0 mL x 0.50 M = 14.0 mmol

excess HCl = 25.0-14.0 = 11.0 mmol
[HCl] excess = 11.0 mmol / (50.0+28.0)ml = 0.141 M

pH = -log_10 (0.141) = 0.85

A.THE PH OF THE SOLUTION LESS THAN 7

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.