During a titration of a weak base with a strong acid, you are slowing converting

Question

During a titration of a weak base with a strong acid, you are slowing converting molecules of the weak base into molecules of its conjugate acid. For the hypothetical weak base, B we see the following: B (aq) + H2O+ (aq) BH+ (aq) + H2O (l) In the problem below you will be adding some strong acid, but not enough to reach the endpoint of the titration. 1.60M 1.80 mL of hydrochloric acid added to the weak base acid in the buret as the titrant 15.0 mL of original 1.50 M NH3 (ammonia) sample You are titrating 15.00 mL of a 1.50 M solution of NH3 (ammonia) with a strong acid. If you add 1.80 mL of a 1.60 M solution of hydrochloric acid, what is the final pH of the remaining weak base solution? The Kb for NH3 (ammonia) is 1.76 X 10-5. final pH =

Explanation / Answer

millimoles of NH3 = 15 x 1.50 = 22.5

millimoles of HCl = 1.80 x 1.60 = 2.88

NH3 + HCl ---------------> NH4Cl

22.5      2.88                         0

19.62     0                         2.88

pKb =-log Kb

pKb = 4.75

poH = pKb + log [NH4Cl / NH3]

pOH = 4.75 + log (2.88/ 19.62)

pOH = 3.92

pH + pOH = 14

pH = 10.08

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

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