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 equivalence point for the titration of 0.10 M KH₂BO₃ with 0.1 M HCl, we can follow these steps:

### Step 1: Understand the Reaction
The reaction occurring during the titration is between the weak acid (boric acid, H₃BO₃) and the strong acid (HCl). At the equivalence point, all the KH₂BO₃ will have reacted with HCl to form H₃BO₃.

### Step 2: Write the Relevant Equilibrium Reaction
The first ionization of boric acid (H₃BO₃) can be represented as:
\[ \text{H}_3\text{BO}_3 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{BO}_3^- + \text{H}_3\text{O}^+ \]

### Step 3: Calculate the Concentration at the Equivalence Point
At the equivalence point, the concentration of H₃BO₃ will be half of the initial concentration of KH₂BO₃, since we are mixing equal volumes of 0.1 M solutions. Therefore, the concentration of H₃BO₃ will be:
\[ \text{Concentration of H}_3\text{BO}_3 = \frac{0.1 \, \text{M} \times V}{V + V} = 0.05 \, \text{M} \]

### Step 4: Set Up the Ka Expression
The acid dissociation constant (Ka) for the first ionization of boric acid is given as:
\[ K_a = \frac{[\text{H}_2\text{BO}_3^-][\text{H}_3\text{O}^+]}{[\text{H}_3\text{BO}_3]} \]
Given \( K_a = 12.8 \times 10^{-10} \), we can substitute the concentrations:
Let \( x \) be the concentration of \( \text{H}_3\text{O}^+ \) at equilibrium. Thus:
\[ K_a = \frac{x^2}{0.05} = 12.8 \times 10^{-10} \]

### Step 5: Solve for x
Rearranging the equation gives:
\[ x^2 = 12.8 \times 10^{-10} \times 0.05 \]
\[ x^2 = 6.4 \times 10^{-11} \]
Taking the square root:
\[ x = \sqrt{6.4 \times 10^{-11}} = 8.0 \times 10^{-6} \]

### Step 6: Calculate the pH
Now, we can find the pH using the concentration of \( \text{H}_3\text{O}^+ \):
\[ \text{pH} = -\log(8.0 \times 10^{-6}) \]
Calculating this gives:
\[ \text{pH} \approx 5.10 \]

### Step 7: Round to the Nearest Whole Number
Rounding 5.10 to the nearest whole number gives us:
\[ \text{pH} \approx 5 \]

### Final Answer
The pH at the equivalence point is **5**.

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