5moldm³ 30cm³ CH₃COOH solution is neutralized by Ba(OH)₂ solution. 0.75dm³ of the solution is used. If 10cm³ of Ba(OH)₂ solution is neutralized by 0.1moldm⁻³ NaOH solution, calculate W.

Chemistry
5moldm³ 30cm³ CH₃COOH solution is neutralized by Ba(OH)₂ solution. 0.75dm³ of the solution is used. If 10cm³ of Ba(OH)₂ solution is neutralized by 0.1moldm⁻³ NaOH solution, calculate W.

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Answer

2.52

Initial amount of CH3COOH\mathrm{CH_3COOH}:
n=0.50 mol dm−3×0.025 dm3=0.0125 moln = 0.50 \, mol \, dm^{-3} \times 0.025 \, dm^3 = 0.0125 \, mol

Equivalence volume of NaOH\mathrm{NaOH}:
Veq=0.0125 mol0.10 mol dm−3=0.125 dm3=125 cm3V_{eq} = \frac{0.0125 \, mol}{0.10 \, mol \, dm^{-3}} = 0.125 \, dm^3 = 125 \, cm^3

Ka=1.8×10−5 mol dm−3K_a = 1.8 \times 10^{-5} \, mol \, dm^{-3}
pKa=−log⁡(1.8×10−5)=4.74\mathrm{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74

(i) Before adding NaOH\mathrm{NaOH}
Weak acid:
[H+]=KaC[H^+] = \sqrt{K_a C}
Substitute:
[H+]=1.8×10−5×0.50=9.0×10−6=3.0×10−3 mol dm−3[H^+] = \sqrt{1.8 \times 10^{-5} \times 0.50} = \sqrt{9.0 \times 10^{-6}} = 3.0 \times 10^{-3} \, mol \, dm^{-3}
pH=−log⁡(3.0×10−3)=2.52pH = -\log(3.0 \times 10^{-3}) = 2.52
2.52

(ii) After adding 10 cm310 \, \mathrm{cm^3} NaOH\mathrm{NaOH}
Moles NaOH\mathrm{NaOH} added:
n=0.10 mol dm−3×0.010 dm3=0.0010 moln = 0.10 \, mol \, dm^{-3} \times 0.010 \, dm^3 = 0.0010 \, mol
Moles CH3COOH\mathrm{CH_3COOH} left:
0.0125−0.0010=0.0115 mol0.0125 - 0.0010 = 0.0115 \, mol
Moles CH3COO−\mathrm{CH_3COO^-}:
0.0010 mol0.0010 \, mol
Total volume:
25+10=35 cm3=0.035 dm325 + 10 = 35 \, cm^3 = 0.035 \, dm^3
Buffer:
pH=pKa+log⁡(nCH3COO−nCH3COOH)pH = pK_a + \log \left( \frac{n_{CH_3COO^-}}{n_{CH_3COOH}} \right)
Ratio:
0.00100.0115=0.087\frac{0.0010}{0.0115} = 0.087
log⁡(0.087)=−1.06\log(0.087) = -1.06
pH=4.74+(−1.06)=3.68pH = 4.74 + (-1.06) = 3.68
3.68

(iii) At equivalence point (125 cm3125 \, \mathrm{cm^3} NaOH\mathrm{NaOH})
Moles CH3COO−=0.0125 mol\mathrm{CH_3COO^-} = 0.0125 \, \mathrm{mol}
Total volume:
25+125=150 cm3=0.150 dm325 + 125 = 150 \, cm^3 = 0.150 \, dm^3
Cb=0.01250.150=0.0833 mol dm−3C_b = \frac{0.0125}{0.150} = 0.0833 \, mol \, dm^{-3}
Kb=1.0×10−141.8×10−5=5.56×10−10 mol dm−3K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} \, mol \, dm^{-3}
Weak base:
[OH−]=KbCb[OH^-] = \sqrt{K_b C_b}
Substitute:
[OH−]=5.56×10−10×0.0833=4.63×10−11=6.8×10−6 mol dm−3[OH^-] = \sqrt{5.56 \times 10^{-10} \times 0.0833} = \sqrt{4.63 \times 10^{-11}} = 6.8 \times 10^{-6} \, mol \, dm^{-3}
pOH=−log⁡(6.8×10−6)=5.17pOH = -\log(6.8 \times 10^{-6}) = 5.17
pH=14.00−5.17=8.83pH = 14.00 - 5.17 = 8.83
8.83

(iv) After adding 30 cm330 \, \mathrm{cm^3} NaOH\mathrm{NaOH}
Moles NaOH\mathrm{NaOH} added:
0.10×0.030=0.0030 mol0.10 \times 0.030 = 0.0030 \, mol
Moles CH3COOH\mathrm{CH_3COOH} left:
0.0125−0.0030=0.0095 mol0.0125 - 0.0030 = 0.0095 \, mol
Moles CH3COO−\mathrm{CH_3COO^-}:
0.0030 mol0.0030 \, mol
Total volume:
25+30=55 cm3=0.055 dm325 + 30 = 55 \, cm^3 = 0.055 \, dm^3
Buffer:
nCH3COO−nCH3COOH=0.00300.0095=0.316\frac{n_{CH_3COO^-}}{n_{CH_3COOH}} = \frac{0.0030}{0.0095} = 0.316
log⁡(0.316)=−0.50\log(0.316) = -0.50
pH=4.74+(−0.50)=4.24pH = 4.74 + (-0.50) = 4.24
4.24

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Initial amount of CH_3COOH: n = 0.50 \, mol \, dm^-3 × 0.025 \, dm^3 = 0.0125 \, mol Equivalence volume of NaOH: V_eq = 0.0125 \, mol0.10 \, mol \, dm^-3 = 0.125 \, dm^3 = 125 \, cm^3 K_a = 1.8 × 10^-5 \, mol \, dm^-3 pK_a = -(1.8 × 10^-5) = 4.74 (i)…

5moldm³ 30cm³ CH₃COOH solution is neutralized by Ba(OH)₂ solution. 0.75dm³ of the solution is used. If 10cm³ of Ba(OH)₂ solution is neutralized by 0.1moldm⁻³ NaOH solution, calculate W.
Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

ScanSolve AI Answer
Initial amount of CH_3COOH: n = 0.50 \, mol \, dm^-3 × 0.025 \, dm^3 = 0.0125 \, mol Equivalence volume of NaOH: V_eq = 0.0125 \, mol0.10 \, mol \, dm^-3 = 0.125 \, dm^3 = 125 \, cm^3 K_a = 1.8 × 10^-5 \, mol \, dm^-3 pK_a = -(1.8 × 10^-5) = 4.74 (i) Before adding NaOH Weak acid: [H^+] = sqrt(K_a C) Substitute: [H^+] = sqrt(1.8 × 10^-5) × 0.50 = sqrt(9.0 × 10^-6) = 3.0 × 10^-3 \, mol \, dm^-3 pH = -(3.0 × 10^-3) = 2.52 2.52 (ii) After adding 10 \, cm^3 NaOH Moles NaOH added: n = 0.10 \, mol \, dm^-3 × 0.010 \, dm^3 = 0.0010 \, mol Moles CH_3COOH left: 0.0125 - 0.0010 = 0.0115 \, mol Moles CH_3COO^-: 0.0010 \, mol Total volume: 25 + 10 = 35 \, cm^3 = 0.035 \, dm^3 Buffer: pH = pK_a + ( n_CH_3COO^-n_CH_3COOH ) Ratio: (0.0010)/(0.0115) = 0.087 (0.087) = -1.06 pH = 4.74 + (-1.06) = 3.68 3.68 (iii) At equivalence point (125 \, cm^3 NaOH) Moles CH_3COO^- = 0.0125 \, mol Total volume: 25 + 125 = 150 \, cm^3 = 0.150 \, dm^3 C_b = (0.0125)/(0.150) = 0.0833 \, mol \, dm^-3 K_b = 1.0 × 10^-141.8 × 10^-5 = 5.56 × 10^-10 \, mol \, dm^-3 Weak base: [OH^-] = sqrt(K_b C_b) Substitute: [OH^-] = sqrt(5.56 × 10^-10) × 0.0833 = sqrt(4.63 × 10^-11) = 6.8 × 10^-6 \, mol \, dm^-3 pOH = -(6.8 × 10^-6) = 5.17 pH = 14.00 - 5.17 = 8.83 8.83 (iv) After adding 30 \, cm^3 NaOH Moles NaOH added: 0.10 × 0.030 = 0.0030 \, mol Moles CH_3COOH left: 0.0125 - 0.0030 = 0.0095 \, mol Moles CH_3COO^-: 0.0030 \, mol Total volume: 25 + 30 = 55 \, cm^3 = 0.055 \, dm^3 Buffer: n_CH_3COO^-n_CH_3COOH = (0.0030)/(0.0095) = 0.316 (0.316) = -0.50 pH = 4.74 + (-0.50) = 4.24 4.24