Question
how would i solve #8?
A 20.0-mL sample of 0.30 M HBr is titrated with 0.15 M NaOH. What is the pH of the solution after 40.3 mL of NaOH have been added to the acid? A) 2.95 B) 3.13 C) 10.87 D) 11.05 E) 13.14 40.0 mL of an acetic acid of unknown concentration is titrated with 0.100 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 5.10. What was the concentration the original acetic acid solution? (K_a(CH_3COOH) = 1.8 times 10^-5) A) 0.11 M B) 0.022 M C) 0.072 D) 0.050 M E) 0.015 M A 25.0-mL sample of 0.10 M C_2H_3NH_2 (ethylamine) is titrated with 0.15 M HCl. What is the pH of solution after 9.00 mL of acid have been added to the amine? K_b = 6.5 times 10^-4 A) 11.08 B) 10.74 C) 10.88 D) 10.55 E) 10.49 When a weak acid is titrated with a strong base, the pH at the equivalence point A) is greater than 7.0 B) is equal to 7.0 C) is less than 7.0 D) is equal to the pK_a of the acid. E) is equal to 14.0 - pk_b, where pk_b is that of the base. When a weak acid is titrated with a weak base, the pH at the equivalence point A) is greater than 7.0. B) is equal to 7.0. C) is less than 7.0. D) is determined by the sizes of K_a and K_b. E) is no longer affected by addition of base.
Explanation / Answer
Q8
V = 40 mL acetic acid, HA
M = 0.1 M of NaOH
V = 20 mL of base, pH= 5.10
find concentration
note that
if pH is stil acidic, this must be in the buffer region
so
pH = pKa + log(A-/HA)
pKa = -log(Ka) = -log(1.8*10^-5) = 4.75
A- = conjguate formed as we added NAOH = Mbase*Vbase = 0.1*20 = 2 mmol
HA = acid left after reaction of NaOH = Macid*Vace - 2 mmol of OH- = 40*Macid - 2
substitute
pH = pKa + log(A-/HA)
5.10 = 4.75 + log(2 / 40*Macid - 2)
10^(5.10-4.75) = 2 / (40*Macid - 2)
(40*Macid - 2) = 2/2.2387
Macid = (2 + 2/2.2387 )/40 = 0.072334 M
choose C
