1) A mixture of equal volumes of equimolar solutions of acetic acid and sodium a
ID: 843810 • Letter: 1
Question
1) A mixture of equal volumes of equimolar solutions of acetic acid and sodium acetate is made in 1 L flask. Calculate the pH of the solution.
2) A sample of 10.0 mL solution of 0. 100 M acetic acid is titrated with 0.100 M NaOH solution. Calculate the pH of the solution when exactly 5.00 mL of the base has been added.
3) Calculate the solubility in molarity of Cu(OH)2,
(a) In pure water and in (b) 0.010M Cu(NO3)2 solution. The Ksp for Cu(OH)2 is 2.2O x 10^-20.
4) (a) Calculate the value for Ksp for CaSO4 that has the solubility of 0.67 g/L in water. (b) If we mix 50.0 mL of 0.100 M Na2SO4 with 5.00 mL 0f 0.100 M BaCl2, will there be any precipitation of BaSO4? Look up for Ksp value for BaSO4.
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Explanation / Answer
1) CH3COOH and CH3COONa forms a acidic buffer system
according to Hasselbach hendersen equation
pH = pka + log [ salt/ acid ]
pH = pKa + log [ CH3COONa / CH3COOH]
given equal volumes of equilmolar solutions
so
[CH3COONa] = [CH3COOH ]
pH = pKa + log [1 ]
pH = pKa
we know that pKa for CH3COOH = 4.76
pH= 4.76
so the pH is 4.76
2)
moles of CH3COOH = molarity x volume /1000
moles of CH3COOH = 0.1 x 10 /1000 = 10 x 10-3
moles of NaOH added = 0.1 x 5 / 1000 = 0.5 x 10-3
CH3COOH + NaOH ----> CH3COONa + H20
from the reaction
moles of CH3OOH reacted = moles of NaOH added = 0.5 x 10-3
moles of CH3COONa formed = moles of CH3COOH reacted = 0.5 x 10-3
moles of CH3COOH remaining = 1 x 10-3 - 0.5 x 10-3 = 0.5 x 10-3
total volume = 10 + 5 = 15 ml
[CH3COOH] = 0.5 x 10-3 x 1000 / 15 = 1/30 M
[CH3COONa] = 0.5 x 10-3 x 1000/ 15 = 1/30 M
pH = pka + log [ salt/ acid ]
pH = pKa + log [ CH3COONa / CH3COOH]
pH = 4.76 + log [ 1/30] / [1/30]
pH = 4.76 + 0
pH = 4.76
so the pH is 4.76
3) Cu[OH]2 ----> Cu+2 + 2OH-
a) in pure water
let the molar solubility be s
Ksp = [Cu+2] [OH-]^2
Ksp = [s] [2s]^2
Ksp = 4s3
2.2 x 10-20 = 4 s3
s = 1.765 x 10-7 M
so the soulbility in water is 1.765 x 10-7 M
b)
in Cu(N03)2
Cu(N03)2 ----> Cu+2 + 2N03-
from the above reaction
[Cu+2] = Cu(N03)2 = 0.01 M
here [Cu+2] is the common ion
so the solubility of Cu(OH)2 depends on [OH-]
now
Ksp = [Cu+2] [OH-]^2
2.2 x 10-20 = 0.01 x [ OH-]^2
[OH-] = 1.48 x 10-9 M
so the solubility is 1.48 x 10-9 M
4) molar mass of CaS04 = 136 g
molar solubility = solubility / molar mass
molar soubility = 0.67 / 136
molar soubility = 4.92 x 10-3
CaS04 ---> Ca+2 + S042-
Ksp = [Ca+2] [S042-]
Ksp = s x s
Ksp = s2
Ksp = ( 4.92 x 10-3 )2
Ksp = 2.427 x 10-5
b) BaS04 ---> Ba+2 + S042-
mol of Ba+2 = 0.1 x 5 /1000 = 5 x 10-4
mol of S042- = 0.1 x 50 /1000 = 5 x 10-3
total volume = 50 + 5 = 55 ml
[Ba+2] = 5 x 10-4 x 1000 / 55 = 9.09 x 10-3
[S042-]= 5 x 10-3 x 1000 /55 = 0.09
Q= [Ba+2] [S042-]
Q = 9.09 x 10-3 x 0.09
Q = 8.26 x 10-4
Ksp of BaS04 = 1.1 x 10-10
as Q> Ksp
Precipitation of BaS04 takes place
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