Consider the titration of 50.5 mL of 1.1 M NaOH with 1.0 M HCl. You may want to

Question

Consider the titration of 50.5 mL of 1.1 M NaOH with 1.0 M HCl. You may want to reference ( pages 682 - 686) Section 16.9 while completing this problem. Part A What is the pH at the start of the titration? Express the pH to two decimal places. Part B What is the pH at the equivalence point? Express the pH to two decimal places. Part C What is the pH after the addition of a large excess of acid (in comparison with the acid volume needed to reach the equivalence point)? Express the pH as an integer.

Explanation / Answer

PART (A)

Intila molarity of NaOH = 1.1 * 50.5 / 1000 = 0.0556 M

So, [NaOH] = [OH-] = 0.0556 M

pOH + pH = 14

pH = 14 - pOH

pH = 14 + Log[OH-]

pH = 14 + Log(0.0556)

pH = 12.7

PART(B)

SInce both NaOH and HCl are strong monoacidic and mono basic solutions respectively. At equivalnece point they form neutral solution whose pH = 7.00

PART(C)

Addition o f large excess of HCl after reaching to equivalence point will increase the concentration of H+ ions which indicates the decrease of pH.

pH of the resulting solution < 7.00. Knowing the exact volume of final solution we can be able to calculate the exact pH.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

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