When 500 , mL of 0.2 , M acetic acid is added | vedclass

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(A) The resulting solution is a buffer solution. The $pH$ is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Aci

Vedclass · Accepted Answer

$5$

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(A) The resulting solution is a buffer solution. The $pH$ is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$ First,calculate $pK_a$: $pK_a = -\log(1.5 	imes 10^{-5}) = 5 - \log(1.5) = 5 - 0.176 = 4.824$ Since the volumes are equal,the ratio of concentrations is equal to the ratio of moles: $pH = 4.824 + \log \left( \frac{0.30}{0.20} ight)$ $pH = 4.824 + \log(1.5) = 4.824 + 0.176 = 5.0$ Thus,the $pH$ of the solution is $5$.

When $500 \, mL$ of $0.2 \, M$ acetic acid is added to $500 \, mL$ of $0.30 \, M$ sodium acetate solution,what is the $pH$ of the resulting solution? (Given: Dissociation constant of acetic acid $K_a = 1.5 \times 10^{-5}$)

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Henderson's equation is $pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}$. If the acid is half-neutralized,the value of $pH$ will be: $[pK_a = 4.30]$

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