Calculate the change in pH when 47.0 mL of a 0.580 M solution of NaOH is added to 1.00 L of a solution that is 1.00 M in sodium acetate and 1.00 M in acetic acid.
for the first ph I get 4.75
then for the ph with naoh I get 4.77
then the difference is .02.
but .02 is incorrect :( thanks for the help
Need help with chemistry problem! thanks :)?
2017-11-06 12:43 pm
回答 (1)
2017-11-06 1:07 pm
Before addition of NaOH :
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) …… Ka
pH before addition of NaOH = pKa + log([CH₃COO⁻]/[CH₃COOH]) = pKa + log(1.00/1.00) = pKa
Initial n(CH₃COO⁻) = Initial n(CH₃COOH) = (1.00 mol/L) × (1.00 L) = 1.00 mol
n(NaOH) added = (0.580 mol/L) × (47/1000 L) = 0.0273 mol
When NaOH is added, each mole of NaOH react with 1 mole of CH₃COOH to form 1 mole of CH₃COO⁻ ions.
After the addition of NaOH :
n(CH₃COOH) = (1.00 - 0.0273) mol = 0.973 mol
n(CH₃COO⁻) = (1.00 + 0.0273) mol = 1.027 mol
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) …… Ka = 1.8 × 10⁻⁵
pH after addition of NaOH = pKa + log([CH₃COO⁻]/[CH₃COOH]) = pKa + log(1.027/0.973)
Change in pH
= (pH after addition of NaOH) - (pH before addition of NaOH)
= log(1.027/0.973)
= 0.0234
≈ 0.0 (to 1 decimal place)
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) …… Ka
pH before addition of NaOH = pKa + log([CH₃COO⁻]/[CH₃COOH]) = pKa + log(1.00/1.00) = pKa
Initial n(CH₃COO⁻) = Initial n(CH₃COOH) = (1.00 mol/L) × (1.00 L) = 1.00 mol
n(NaOH) added = (0.580 mol/L) × (47/1000 L) = 0.0273 mol
When NaOH is added, each mole of NaOH react with 1 mole of CH₃COOH to form 1 mole of CH₃COO⁻ ions.
After the addition of NaOH :
n(CH₃COOH) = (1.00 - 0.0273) mol = 0.973 mol
n(CH₃COO⁻) = (1.00 + 0.0273) mol = 1.027 mol
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) …… Ka = 1.8 × 10⁻⁵
pH after addition of NaOH = pKa + log([CH₃COO⁻]/[CH₃COOH]) = pKa + log(1.027/0.973)
Change in pH
= (pH after addition of NaOH) - (pH before addition of NaOH)
= log(1.027/0.973)
= 0.0234
≈ 0.0 (to 1 decimal place)
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