Question

The equilibrium dissociation constant of acetic acid is \( 1.74 \times 10^{-5} \) M. The \( pK_a \) of acetic acid (rounded off to one decimal place) is ......

Hint:

To calculate the \( pK_a \) of a weak acid, use the formula \( pK_a = -\log K_a \). The smaller the \( K_a \), the weaker the acid, and the larger the \( pK_a \).

Answer 1

Correct Answer: 4.7

To calculate the \( pK_a \) of acetic acid, we use the formula: \[ pK_a = -\log K_a \] where \( K_a \) is the equilibrium dissociation constant of acetic acid. Given that \( K_a = 1.74 \times 10^{-5} \), we can substitute this value into the equation: \[ pK_a = -\log (1.74 \times 10^{-5}) \] Now, applying the logarithmic properties: \[ pK_a = -\log (1.74) - \log (10^{-5}) \] We know that \( \log (10^{-5}) = -5 \), so: \[ pK_a = -\log (1.74) + 5 \] Using the value \( \log (1.74) \approx 0.240 \), we get: \[ pK_a = -0.240 + 5 = 4.760 \] Rounding this value to one decimal place, we obtain: \[ pK_a \approx 4.8 \] Thus, the \( pK_a \) of acetic acid is approximately 4.8.