2. On the basis of your results, what is the relationship between the temperature of the solvent and the rate of solution format
1 answer:
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Question options:
a) 2.05
b) 0.963
c) 0.955
d) 1.00
Answer:
b) 0.963
Explanation:
H2SO4→ HSO4- + H3O+
HSO4- + H2O ⇌ SO42- + H3O+
Construct ICE table:
HSO4- (aq) + H2O ⇌ SO42- (aq) + H3O+ (aq)
I 0.1 solid & 0 0.1
C -x liquid + x + x
E 0.1 - x are ignored x 0.1 + x
Calculate x
Ka = products/reactants
=
0.011 =
0.011 x (0.1 -x) = o.1x + x^2
0.0011 - 0.011 x - o.1x - x^2 = 0
0.0011 - 0.011 x - x^2 = 0
Use formula to solve for quadratic equation
x = / 2a
a = -1, b = -0.111, c = 0.001
Solve for x
x = / 2(-1)
x = 0.111 +,- / -2
x = 0.111 +,- / -2
x =
x = , x =
x = , x =
x = - 0.12015 , x = 0.00915
x cannot be negative, so
x = 0.00915 M
Calculate [H3O+]
[H3O+] = 0.1 M + x
[H3O+] = 0.1 M + 0.00915 M
[H3O+] = 0.10915 M
Clculate pH
pH = - log [ H3O+]
pH = - log [ 0.10915]
pH = 0.963
Q1)
This is a strong acid- strong base base reaction, balanced equation for the reaction is as follows;
LiOH + HCl ---> LiCl + H₂O
stoichiometry of acid to base is 1:1
volume of HCl used up - 13.60 - 1.25 = 12.35 mL
volume of LiOH used up - 11.20 - 2.65 = 8.55 mL
molarity of LiOH - 0.140 M
The number of LiOH moles reacted -
= 0.001197 mol
according to stoichiometry, number of LiOH moles = number of HCl moles
Therefore number of HCl moles reacted - 0.001197 mol
The number of HCl moles in 12.35 mL - 0.001197 mol
Then number of HCl moles in 1000 mL -
Molarity of HCl - 0.0969 M
Q2)
Volume of HCl used - 12.35 mL
Volume of LiOH used - 8.55 mL
Molarity of HCl - 0.140 M
In 1 L solution of HCl there are 0.140 mol of HCl
Therefore number of HCl moles in 12.35 mL -
Number of HCl moles reacted - 0.001729 mol
since molar ratio of acid to base is 1:1
the number of LiOH moles that reacted - 0.001729 mol
Therefore number of moles in 8.55 mL - 0.001729
Then number of LiOH moles in 1000 mL -
molarity of LiOH - 0.202 M
This is a strong acid- strong base base reaction, balanced equation for the reaction is as follows;
LiOH + HCl ---> LiCl + H₂O
stoichiometry of acid to base is 1:1
volume of HCl used up - 13.60 - 1.25 = 12.35 mL
volume of LiOH used up - 11.20 - 2.65 = 8.55 mL
molarity of LiOH - 0.140 M
The number of LiOH moles reacted -
according to stoichiometry, number of LiOH moles = number of HCl moles
Therefore number of HCl moles reacted - 0.001197 mol
The number of HCl moles in 12.35 mL - 0.001197 mol
Then number of HCl moles in 1000 mL -
Molarity of HCl - 0.0969 M
Q2)
Volume of HCl used - 12.35 mL
Volume of LiOH used - 8.55 mL
Molarity of HCl - 0.140 M
In 1 L solution of HCl there are 0.140 mol of HCl
Therefore number of HCl moles in 12.35 mL -
Number of HCl moles reacted - 0.001729 mol
since molar ratio of acid to base is 1:1
the number of LiOH moles that reacted - 0.001729 mol
Therefore number of moles in 8.55 mL - 0.001729
Then number of LiOH moles in 1000 mL -
molarity of LiOH - 0.202 M