Chapter 8: Problem 164
\(18 \mathrm{ml}\) of mixture of acetic acid and sodium acetate required \(6 \mathrm{ml}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\) for neutralization of the acid and \(12 \mathrm{ml}\) of \(0.1 \mathrm{M} \mathrm{HCl}\) for reaction with salt separately. If \(\mathrm{pK}_{\mathrm{a}}\) of the acid is \(4.75\), what is the \(\mathrm{pH}\) of the mixture? (a) \(5.05\) (b) \(4.75\) (c) \(4.5\) (d) \(4.6\)
Short Answer
Step by step solution
Determine Moles of NaOH Required
Calculate Moles of HCl Required
Apply Henderson-Hasselbalch Equation
Calculate Logarithmic Value
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Henderson-Hasselbalch Equation
- \(\mathrm{pH}\) is what we want to find.
- \(\mathrm{pK}_a\) represents the strength of the acid, with smaller values indicating stronger acids.
- \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the undissociated weak acid.
Calculating pH
- From the neutralization process with NaOH, you find the moles of acetic acid, which got neutralized to form acetate ions.
- From the acidic reaction with HCl, additional moles of acetate are accounted for by releasing acetic acid back into the mixture.
Neutralization Reaction
Buffer Solutions
- If acid is added (in the form of HCl), the acetate ions react to form acetic acid, thereby minimizing pH change.
- If a base like NaOH is added, it reacts with the acetic acid to form more acetate, stabilizing the pH.