A) A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated
with 0.40 MM NaOH. Calculate the pH after the addition of 17.0 mL
of NaOH.
B) Calculate the molar solubility of lead thiocyanate in 1.00 M
KSCN. Lead thiocyanate, Pb(SCN)2 has a Ksp value of 2.00×10−5.
C) Calculate the molar solubility of AgBr in 0.10 M NaBr
solution.
D) Calculate the molar solubility of Ni(OH)2Ni(OH)2 when
buffered at pH=pH= 10.2. The Ksp of nickel hydroxide =6.0×10−16
M
imagine mixing 25.0 ml of 0.10 M acetic acid (Ka=1.8*10^-5) with
25.0 ml of 0.10 M sodium acetate and determine a) The initial pH of
the buffer b) pH of 20.0 ml sample of buffer with 1.0 ml of 0.10M
Hal added c) pH of another 20 mL sample of buffer with 1.0 ml of
0.10 M MaOH added
Consider the titration of a 30.0 mL sample of 0.150 M CH3COOH
(Ka=1.8×10−5) with 0.200 M NaOH. Determine each quantity: Part A:
What is the volume of base required to reach the equivalence point?
Express your answer using one decimal place. Part B: What is the pH
after 10.0 mL of base have been added? Express your answer using
two decimal places. Part C What is the pH at the equivalence point?
Express your answer using two decimal places. Part...
Part C
Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8×10−5)
with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the
titrant at 25 ∘C.
Part D
A 30.0-mL volume of 0.50 M CH3COOH
(Ka=1.8×10−5) was titrated with 0.50 M NaOH.
Calculate the pH after addition of 30.0 mL of NaOH at 25 ∘C.
at 25 C, a solution is prepared by adding 50 ml of 0.2 M NaOH to
75 ml of 0.1 M NaOH. What is the total concentratoin of hydorxide
ion in the solution? Assume volumes are additive.
Consider the titration of 10.00 mL of 0.10 M acetic acid
(CH3COOH) with 0.10 M NaOH
a. What salt is formed during this reaction?
b. do you expect the salt solution at the equivalence point to
be acidic, neutral, or basic?
Calculate the pH of this solution at the equicalence point.
Calculate the pH of the solution resulting from the addition of
10.0 mL of 0.10 M NaOH to 50.0 mL of a 0.10 M solution of aspirin
(acetylsalicic acid, Ka = 3.0 × 10–4)
solution.
A.
2.9
B.
10.5
C.
4.1
D.
3.5
E.
1.8