The equilibrium constant for the following reaction in aqueous solution is `0.90.`
`H_(3)BO_(3)+"glycerin"hArr(H_(3)BO_(3)-"glycerin")`
How many mole of glycerin should be added per litre of `0.10 M H_(3)BO_(3)` so that `80%` of the `H_(3)BO_(3)` is converted to the boric-acid glycerin complex ?

Options

A. `4.44`
B. `4.52`
C. `3.6`
D. `0.08`

Answer

To solve the problem step by step, we will analyze the given reaction and the equilibrium constant, and then calculate the amount of glycerin needed.

### Step 1: Write the Reaction and Equilibrium Expression
The reaction is:
\[ \text{H}_3\text{BO}_3 + \text{glycerin} \rightleftharpoons \text{H}_3\text{BO}_3\text{-glycerin} \]

The equilibrium constant expression \( K_c \) for this reaction is given by:
\[ K_c = \frac{[\text{H}_3\text{BO}_3\text{-glycerin}]}{[\text{H}_3\text{BO}_3][\text{glycerin}]} \]

Given that \( K_c = 0.90 \).

### Step 2: Initial Concentrations
We start with a concentration of \( \text{H}_3\text{BO}_3 \) of \( 0.10 \, \text{M} \). Let the initial concentration of glycerin be \( A \) (in M).

### Step 3: Change in Concentrations
If 80% of \( \text{H}_3\text{BO}_3 \) reacts, then:
\[ \text{Amount of } \text{H}_3\text{BO}_3 \text{ reacted} = 0.80 \times 0.10 \, \text{M} = 0.08 \, \text{M} \]

At equilibrium:
- Concentration of \( \text{H}_3\text{BO}_3 = 0.10 - 0.08 = 0.02 \, \text{M} \)
- Concentration of \( \text{H}_3\text{BO}_3\text{-glycerin} = 0.08 \, \text{M} \)
- Concentration of glycerin = \( A - 0.08 \)

### Step 4: Substitute into Equilibrium Expression
Substituting the equilibrium concentrations into the \( K_c \) expression:
\[ 0.90 = \frac{0.08}{(0.02)(A - 0.08)} \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ 0.90(A - 0.08) = 0.08 / 0.02 \]
\[ 0.90(A - 0.08) = 4 \]

### Step 6: Solve for A
Now, solving for \( A \):
\[ 0.90A - 0.072 = 4 \]
\[ 0.90A = 4 + 0.072 \]
\[ 0.90A = 4.072 \]
\[ A = \frac{4.072}{0.90} \]
\[ A \approx 4.52 \, \text{M} \]

### Conclusion
Thus, the amount of glycerin that should be added per liter of \( 0.10 \, \text{M} \, \text{H}_3\text{BO}_3 \) is approximately \( 4.52 \, \text{M} \).

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