Weak acids and bases are those that do not completely dissociate in water. the dissociation of the acid or base is an equilibrium process and has a corresponding equilibrium constant. ka is the equilibrium constant for the dissociation of a weak acid and kb is the equilibrium constant for the dissociation of a weak base. what is the ph of a solution that has 0.125 m ch3cooh and 0.125 m h3bo3? ka of ch3cooh = 1.8 × 10−5 and ka of h3bo3 = 5.4 × 10−10 answer unselected 4.74 unselected 5.64 unselected 2.82 unselected 2.52 unselected

pH=2.82

Explanation:

pH is a scale used to indicate if a water-based solution is acidic or basic, and is defined by the following equation:

 Equation (1)

In aqueous solutions there are no ions , because the proton is transferred to to form hydronium ions, . Thus, we can determine the pH of the solution, finding the molar concentration of ions:

Equation (2)

The solution prepared contains two weak acids, each one undergoes an equilibrium process with the corresponding equilibrium constant Ka, as it is represented in the following reaction for a weak acid HA:

HA  +    ⇔    +  

The equilibrium constant is defined by the equilibrium concentration of products over reactants:

 Equation (3)

However, the molar concentration of water is essentially constant for reactions in aqueous solutions, then the acid dissociation constant is defined as follow:

 Equation (4)

Information for :

Reaction:   + ⇔  +    

Initial moles per Liter:         0.125 M       +     A        ⇔            0        +       0

Reacting moles per liter:        -x          +      -x            ⇔             x       +     x

Equilibrium mole per liter:  0.125 M-x+      A-x         ⇔            x      +      x

Information for :

Reaction:     + ⇔ +

Initial moles per Liter:        0.125 M     +        A      ⇔               0        +       0

Reacting moles per liter:        -y            +       -y          ⇔               y       +     y

Equilibrium mole per liter:   0.125 M-y+       A-y        ⇔              y      +      y

Replacing the equilibrium information of each acid in equation 4, we get:

For :

Equation (5)

For :

Equation (6)

Solving equation 5 and 6:

For :

For :

Due to both acids are in the same solution, the total concentration of is (x+y) . Replacing this concentration in equation 2, we get:

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol