At `90^(@)`C, pure water has `[H^(+)]=10^(-6)` M.If 100 mL of `0.2` M HCl is added to 200 mL of `0.1` M KOH at `90^(@)`C then pH of the resulting solution will be :
A
5
B
6
C
7
D
None of these
Text Solution
AI Generated Solution
The correct Answer is:
To find the pH of the resulting solution when 100 mL of 0.2 M HCl is added to 200 mL of 0.1 M KOH at 90°C, we can follow these steps:
### Step 1: Calculate the concentration of H⁺ ions in pure water at 90°C.
At 90°C, the concentration of H⁺ ions in pure water is given as [H⁺] = 10⁻⁶ M.
### Step 2: Calculate the number of milliequivalents of HCl.
To find the number of milliequivalents of HCl, we use the formula:
\[
\text{Milliequivalents of HCl} = \text{Volume (mL)} \times \text{Concentration (M)}
\]
For HCl:
\[
\text{Milliequivalents of HCl} = 100 \, \text{mL} \times 0.2 \, \text{M} = 20 \, \text{meq}
\]
### Step 3: Calculate the number of milliequivalents of KOH.
Similarly, for KOH:
\[
\text{Milliequivalents of KOH} = 200 \, \text{mL} \times 0.1 \, \text{M} = 20 \, \text{meq}
\]
### Step 4: Determine the neutralization reaction.
When HCl (which provides H⁺ ions) is added to KOH (which provides OH⁻ ions), they will neutralize each other:
\[
\text{H⁺} + \text{OH⁻} \rightarrow \text{H₂O}
\]
Since both HCl and KOH provide 20 milliequivalents, they will completely neutralize each other.
### Step 5: Determine the pH of the resulting solution.
After complete neutralization, the solution will be neutral. At 90°C, the pH of pure water is 6 (as given by the concentration of H⁺ ions). Therefore, the pH of the resulting solution will also be 6.
### Final Answer:
The pH of the resulting solution will be **6**.
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