Given the following titration curve answer questions 4-6. Make sure your mark on

Question

Given the following titration curve answer questions 4-6. Make sure your mark on the graph the volume at the equivalence point, the pH at the equivalence point, the volume where the pH-pK, and the value for the pK 15.00 14.00 13.00 12.00 11.00 10.00 9.00 8.00 pH 7.00 6.00 5.00 4.00 3.00 2.00 1.00 0.00 0 48 12 16 20 24 28 32 36 40 4448 52 56 60 Volume of Titrant (mL) The above titration curve was obtained when a 10.00 mL sample of a 0.50 M base was titrated with an acid. What is the approximate molarity of the acid (the titrant) used? 4. 5. What is the value of the Kb for the base? (Hint: see p. 603 Section 15.8 of Tro.) 6. What is the pH at the equivalence point? Why is the pH not-7.00? Explain

Explanation / Answer

Q4

Molarity of acid is given in equivalence point

mol of aicd = mol of base

Vequicalence = 40 mL approx...

mol of base = MV = 0.5 * 10 = 5mol = 5*10 ^-3 mol

mol of acid = mol of base = 5*10^-3

then

[HA] = (5*10^-3) / (40*10^-3) = 0.125 M

Q5.

for Kb of the base...

at half equivalence point

pH = pKa + log(A-/HA)

A- = HA

pH = pKa + log(1) in half equivlanec epoint

pH = pKa

Vhalf = 1/2*Ve2 = 40/2 = 20 mL

pH at V = 20 mL = 10 approx

pOH = 14-pH = 14-10 = 4

pKb = pOH = 4

Kb = 10^-4

Q6.

pH in equivalence point is approx

pH = 6

it is not 7 since it is a WEAK base, it will form hydrolysis

BH+ + H2O <--> B + H3O+

there are H3O+ in equilbirium, so it will be more acidic

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

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