QQuestionChemistry
QuestionChemistry
A 30.0 mL volume of 0.50 M CH^3COOH (Ka= 1.8 * 10^- 5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH.
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Answer
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Step 1:Let's solve this acid-base titration problem step by step:
Step 2:: Determine the initial moles of acetic acid (CH₃COOH)
n_{CH_{3}COOH} = (0.030 \mathrm{L}) \times (0.50 \mathrm{M}) = 0.015 \mathrm{mol}
Step 3:: Determine the moles of NaOH added
n_{NaOH} = (0.030 \mathrm{L}) \times (0.50 \mathrm{M}) = 0.015 \mathrm{mol}
Step 4:: Analyze the reaction
CH₃COOH + NaOH → CH₃COONa + H₂O - Moles of acid and base are equal - This means we're at the equivalence point - We've formed a salt of a weak acid (sodium acetate)
Step 5:: Calculate the salt concentration
- Concentration of CH₃COONa = $$\frac{0.015 \mathrm{mol}}{0.060 \mathrm{L}} = 0.25 \mathrm{M}
- Total volume = 60.0 mL = 0.060 L
Step 6:: Calculate pH using the hydrolysis of the salt
- Use the Ka of acetic acid: $$K_{a} = 1.8 \times 10^{-5}
- Salt hydrolysis: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
Step 7:: Calculate Kb for acetate
K_{b} = \frac{K_{w}}{K_{a}} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}
Step 8:: Calculate pOH using the Kb expression
pOH = -\log\left(\sqrt{\frac{K_{b}}{C_{salt}}}\right)
pOH = -\log\left(\sqrt{\frac{5.56 \times 10^{- 10}}{0.25}}\right) = 4.62
Step 9:: Calculate pH
pH = 14 - pOH = 14 - 4.62 = 9.38
Final Answer
The pH after adding 30.0 mL of NaOH is 9.38.
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