Acid-base indicators are either weak organic acids or weak organic bases. Indicator change colour in dilute solution when the hydonium ion concentration reaches a particular calur For example. Phenolphthalein is a coloureless stbstance in any aqueous solution with a pH less than 8.3 In between the pH range 8.3 to 10, transition of colour (colourless to pink ) takes place and if pH of solution is greater than 10 solution is dark pink. Considering an acid indicator Hln, the equilibrium involving it and its conjgate base `In^(-)` can be represented as :
`" " underset("acidic from")(HIn)hArrH^(+)underset("basic from")(In^(-))`
pH of solution can be computed as :
`" " pH=pK_(In)+log.([IN^(-)])/([HIn])`
In general, transition of colour takes place in between the pH range `pK_(In+-1.`
Calculate the pH at equivalence point when 5 milli mol of HB is titrated with 0.1 M NaOH.

Options

A. 8.75
B. 8.85
C. `9.0`
D. None of these

Answer

To solve the problem of calculating the pH at the equivalence point when 5 millimoles of HB are titrated with 0.1 M NaOH, we can follow these steps:

### Step 1: Determine the initial moles of the acid (HB)
Given that we have 5 millimoles of HB, we can convert this to moles:
\[ \text{Moles of HB} = 5 \text{ millimoles} = 0.005 \text{ moles} \]

### Step 2: Calculate the volume of NaOH needed to reach the equivalence point
At the equivalence point, the moles of NaOH will equal the moles of HB. Since the concentration of NaOH is 0.1 M:
\[ \text{Volume of NaOH (L)} = \frac{\text{Moles of HB}}{\text{Concentration of NaOH}} \]
\[ \text{Volume of NaOH (L)} = \frac{0.005 \text{ moles}}{0.1 \text{ M}} = 0.05 \text{ L} = 50 \text{ mL} \]

### Step 3: Calculate the total volume of the solution at the equivalence point
The total volume of the solution after adding 50 mL of NaOH to the initial volume of HB (assuming it started with negligible volume) will be:
\[ \text{Total Volume} = 50 \text{ mL} + 0 \text{ mL} = 50 \text{ mL} \]

### Step 4: Determine the concentration of the conjugate base (A^-)
At the equivalence point, all of the acid (HB) has been converted to its conjugate base (B^-). The concentration of B^- can be calculated as:
\[ \text{Concentration of B^-} = \frac{\text{Moles of B^-}}{\text{Total Volume (L)}} \]
\[ \text{Concentration of B^-} = \frac{0.005 \text{ moles}}{0.050 \text{ L}} = 0.1 \text{ M} \]

### Step 5: Use the Kb to find the pH
The dissociation constant \( K_a \) for HB is given as \( 10^{-5} \). We can find \( K_b \) using the relation:
\[ K_w = K_a \cdot K_b \]
Where \( K_w = 10^{-14} \):
\[ K_b = \frac{K_w}{K_a} = \frac{10^{-14}}{10^{-5}} = 10^{-9} \]

### Step 6: Set up the equilibrium expression for B^-
The dissociation of B^- can be represented as:
\[ B^- + H_2O \rightleftharpoons HB + OH^- \]
Using the expression for \( K_b \):
\[ K_b = \frac{[HB][OH^-]}{[B^-]} \]
Let \( x \) be the concentration of \( OH^- \) produced:
\[ K_b = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1} \]
Since \( K_b \) is small, we can ignore \( x \) in the denominator:
\[ 10^{-9} = \frac{x^2}{0.1} \]
\[ x^2 = 10^{-10} \]
\[ x = 10^{-5} \text{ M} \]

### Step 7: Calculate the pOH and then pH
\[ pOH = -\log(10^{-5}) = 5 \]
\[ pH = 14 - pOH = 14 - 5 = 9 \]

### Final Answer
The pH at the equivalence point is **9**.