The pH at the equivalent point for the titration of 0.10 M `KH_(2)BO_(3)` with 0.1 M HCl is `(K_(a)" of "H_(3)BO_(3)=12.8xx10^(-10))`
Report your answer by rounding it up to nearest whole number.

Answer

To find the pH at the equivalent point for the titration of 0.10 M KH₂BO₃ with 0.1 M HCl, we follow these steps:

### Step 1: Understand the Reaction
The reaction between potassium borate (KH₂BO₃) and hydrochloric acid (HCl) produces boric acid (H₃BO₃) and potassium chloride (KCl). The balanced equation is:
\[ \text{KH}_2\text{BO}_3 + \text{HCl} \rightarrow \text{H}_3\text{BO}_3 + \text{KCl} \]

### Step 2: Determine the Concentration of H₃BO₃ at the Equivalent Point
At the equivalent point, all of the KH₂BO₃ has reacted with HCl to form H₃BO₃. Since both solutions are 0.1 M and assuming equal volumes, the concentration of H₃BO₃ will also be 0.1 M.

### Step 3: Calculate the Ionization of H₃BO₃
H₃BO₃ is a weak acid, and we need to find its dissociation constant (Kₐ) to calculate the concentration of hydrogen ions (H⁺) in solution. The given \( K_a \) for H₃BO₃ is \( 12.8 \times 10^{-10} \).

### Step 4: Use the Ionization Formula
The ionization of H₃BO₃ can be represented as:
\[ \text{H}_3\text{BO}_3 \rightleftharpoons \text{H}^+ + \text{H}_2\text{BO}_3^- \]

Using the formula for the degree of ionization (α):
\[ K_a = \frac{[H^+][H_2BO_3^-]}{[H_3BO_3]} \]
Assuming that x is the concentration of \( H^+ \) ions produced, we can write:
\[ K_a = \frac{x^2}{0.1 - x} \]

For weak acids, if \( K_a \) is small compared to the concentration, we can approximate:
\[ K_a \approx \frac{x^2}{0.1} \]

### Step 5: Solve for x
Substituting the value of \( K_a \):
\[ 12.8 \times 10^{-10} = \frac{x^2}{0.1} \]
\[ x^2 = 12.8 \times 10^{-10} \times 0.1 \]
\[ x^2 = 1.28 \times 10^{-11} \]
\[ x = \sqrt{1.28 \times 10^{-11}} \]
\[ x \approx 3.58 \times 10^{-6} \]

### Step 6: Calculate the pH
The concentration of \( H^+ \) ions is approximately \( 3.58 \times 10^{-6} \). The pH is calculated using:
\[ \text{pH} = -\log[H^+] \]
\[ \text{pH} = -\log(3.58 \times 10^{-6}) \]
Using logarithmic properties:
\[ \text{pH} \approx 5.45 \]

### Step 7: Round to the Nearest Whole Number
Rounding 5.45 to the nearest whole number gives us:
\[ \text{pH} \approx 5 \]

### Final Answer
The pH at the equivalent point for the titration is **5**.

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