To determine the pH of a solution which has 0.195 M hc2h3o2 and 0.125 M kc2h3o2, we use the ICE table and the acid dissociation constant of hc2h3o2 <span>to determine the concentration of the hydrogen ion present at equilibrium. We do as follows:
HC2H3OO = H+ + </span>C2H3OO-
KC2H3OO = K+ + C2H3OO-
Therefore, the only source of hydrogen ion would be the acid. We use the ICE table,
HC2H3OO H+ C2H3OO-
I 0.195 0 0.125
C -x +x +x
------------------------------------------------------------------
E 0.195-x x 0.125 + x
Ka = <span>1.8*10^-5 = (0.125 + x) (x) / 0.195 -x
x = 2.81x10^-5 M = [H+]
pH = - log [H+]
pH = -log 2.81x10^-5
pH = 4.55
Therefore, the pH of the resulting solution would be 4.55.</span>
Let's start to understand this question by a simple combustion reaction involving oxidation of Ethane in the presence of Oxygen. When Ethane is burned in the presence of Oxygen it produces Carbon Dioxide and Water respectively. Therefore, the equation is as,
C₂H₆ + O₂ → CO₂ + H₂O
Above reaction shows the reaction and the equation is unbalanced. Balancing chemical equation is important because according to law of conservation of mass, mass can neither be created nor destroyed. Hence, we should balance the number of elements on both side.
LHS RHS
Carbon Atoms 2 1
Hydrogen Atoms 6 2
Oxygen Atoms 2 3
It means this equation is not obeying the law. Now, how to balance? One way is as follow,
C₂H₆ + O₃ → C₂O₂ + H₆O
LHS RHS
Carbon Atoms 2 2
Hydrogen Atoms 6 6
Oxygen Atoms 3 3
We have balanced the equation by changing the subscripts. But, we have messed up the chemical composition of compounds and molecules like Oxygen is converted into Ozone.
Therefore, we will change the coefficients (moles) to balance the equation as,
C₂H₆ + 7/2 O₂ → 2 CO₂ + 3 H₂O
LHS RHS
Carbon Atoms 2 2
Hydrogen Atoms 6 6
Oxygen Atoms 7 7
Now, by changing the coefficients we have balanced the equation without disturbing the chemical composition of compounds and molecules.
Answer:
Option 4
Explanation:
Velocity is the rate at which the position changes
When a mixture of 10 moles of SO2 and 15 moles of O2 was passed over a catalyst, 10 moles of SO3 was formed.
Answer: Thus the cell potential of an electrochemical cell is +0.28 V
Explanation:
The calculation of cell potential is done by :
Where both are standard reduction potentials.
As Reduction takes place easily if the standard reduction potential is higher(positive) and oxidation takes place easily if the standard reduction potential is less(more negative). Thus iron acts as anode and lead acts as cathode.
Thus the cell potential of an electrochemical cell is +0.28 V