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Answer :
Final answer:
The initial pH of a solution of acetic acid can be determined by taking the negative logarithm of the hydronium ion concentration, which is calculated using the acid dissociation constant (Ka) and the concentration of the acetic acid. After the solving the equation the correct answer is option b) 1.70.
Explanation:
The constant for \( \text{HClO} \) is the acid dissociation constant (\( K_a \)).
The \( K_a \) for \( \text{HClO} \) (chlorous acid) is approximately \( 1.1 \times 10^{-2} \).
Before the addition of any \( \text{KOH} \) (potassium hydroxide), the pH of a solution containing only \( \text{HClO} \) can be calculated using the expression for the acid dissociation constant (\( K_a \)):
\[ K_a = \frac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]} \]
Given that \( K_a = 1.1 \times 10^{-2} \), the pH before the addition of any \( \text{KOH} \) can be determined using the \( K_a \) expression and solving for \( \text{H}^+ \) concentration.
Since the initial concentration of \( \text{HClO} \) is the same as \( [\text{H}^+] \) initially, and \( [\text{H}^+] = [\text{ClO}^-] \) due to the 1:1 stoichiometry of the dissociation, we have:
\[ K_a = \frac{[\text{H}^+]^2}{[\text{HClO}]} \]
\[ 1.1 \times 10^{-2} = \frac{x^2}{0.010 \, \text{M}} \]
Solving for \( x \), we find:
\[ x = [\text{H}^+] = 0.105 \, \text{M} \]
Thus, the pH before the addition of any \( \text{KOH} \) is approximately \( -\log(0.105) \approx 1.70 \).
So, the correct answer is option (b) 1.70.
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