10. 0.01 mol of HCl were added to 1 Lofa buffer solution containing 0.08 mol of CH_(3)COOH and 0.04 mol of CH_(3)COONa. Calculate the pH of the resulting solution (forCH_(3)COOHpK_(a)=4.76) . (round up the number to the nearest tenth) square

To calculate the pH of the resulting solution, we can use the Henderson-Hasselbalch equation, which is given by:<br /><br />\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]<br /><br />where:<br />- \(\text{p}K_a\) is the negative logarithm of the acid dissociation constant (\(K_a\)) of the acid.<br />- \([\text{Base}]\) is the concentration of the conjugate base.<br />- \([\text{Acid}]\) is the concentration of the acid.<br /><br />Given:<br />- \(\text{p}K_a\) for \(CH_3COOH\) = 4.76<br />- Initial concentration of \(CH_3COOH\) = 0.08 mol<br />- Initial concentration of \(CH_3COONa\) (conjugate base) = 0.04 mol<br />- Amount of \(HCl\) added = 0.01 mol<br /><br />First, we need to determine the new concentrations of \(CH_3COOH\) and \(CH_3COONa\) after adding \(HCl\).<br /><br />Since \(HCl\) is a strong acid, it will dissociate completely in water, providing \(H^+\) ions. The \(H^+\) ions will react with the conjugate base (\(CH_3COONa\)) to form more \(CH_3COOH\):<br /><br />\[ H^+ + CH_3COO^- \rightarrow CH_3COOH \]<br /><br />The new concentrations will be:<br />- New concentration of \(CH_3COOH\) = Initial concentration of \(CH_3COOH\) + amount of \(HCl\) added<br /> \[ 0.08 \, \text{mol} + 0.01 \, \text{mol} = 0.09 \, \text{mol} \]<br />- New concentration of \(CH_3COONa\) = Initial concentration of \(CH_3COONa\) - amount of \(HCl\) added<br /> \[ 0.04 \, \text{mol} - 0.01 \, \text{mol} = 0.03 \, \text{mol} \]<br /><br />Now, we can plug these values into the Henderson-Hasselbalch equation:<br /><br />\[ \text{pH} = 4.76 + \log \left( \frac{0.03}{0.09} \right) \]<br /><br />Calculate the ratio inside the logarithm:<br /><br />\[ \frac{0.03}{0.09} = \frac{1}{3} \approx 0.333 \]<br /><br />Now, calculate the logarithm:<br /><br />\[ \log (0.333) \approx -0.477 \]<br /><br />Finally, add this value to the \(\text{p}K_a\):<br /><br />\[ \text{pH} = 4.76 + (-0.477) \approx 4.76 - 0.477 = 4.283 \]<br /><br />Rounding to the nearest tenth:<br /><br />\[ \text{pH} \approx 4.3 \]<br /><br />So, the pH of the resulting solution is approximately 4.3.