A 100 cm^3 sample of sodium hydroxide ....?

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question ) A 100 cm^3 sample of sodium hydroxide of concentration 0.600 mol dm^-3 is mixed with 200 cm^3 of ethanoic acid of concentration 0.600 mol dm^-3

1) write an equation for the reaction which occurs?

2) calculate the number of moles unreacted ethanoic acid remaining in the resulting mixture?

3) calculate the concentration, in mol dm^-3 of the unreacted ethanoic acid in the resulting mixture?

4) the mixture which forms is a buffer. why does the pH of the mixture remain constant when small quantities of solutions containing H+ of OH- ions are added? ( use equations in your explanation.)

5) the concentration of sodium ethanoate in the mixture is 0.200 mol dm^-3. calculate the pH of the mixture, using your answer to 3

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  • 8 years ago
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    question ) A 100 cm^3 sample of sodium hydroxide of concentration 0.600 mol dm^-3 is mixed with 200 cm^3 of ethanoic acid of concentration 0.600 mol dm^-3

    ethanoic acid = CH3COOH When reacted with NaOH, the H+ bonds with OH- to produce H2O. The Na+1 and CH3COO-1 form NaCH3COOH and H2O. CH3COOH is a weak acid in vinegar, and NaCH3COOH is the salt of the weak acid.

    1) write an equation for the reaction which occurs?

    NaOH + CH3COOH → NaCH3COO + H2O

    2) calculate the number of moles unreacted ethanoic acid remaining in the resulting mixture?

    According the balanced equation, 1 mole of NaOH reacts with 1 mole of CH3COOH to produce 1 mole of NaCH3COO and 1 mole of H2O.

    Determine the number of moles of each reactant.

    Number of moles = molarity * liters

    1 cm^3 = 1 ml

    1 dm^3 = 10^3 cm^3 = 1000 cm^3 = 1 liter

    100 cm^3 = 100 ml = 0.100 Liter

    0.600 mol dm^-3 = 0.600 mole/Liter

    Number of moles of NaOH = 0.600 mole/liter * 0.100 liter = 0.06 moles of NaOH

    Number of moles of CH3COOH = 0.600 mole/liter * 0.200 liter = 0.12 moles of NaOH

    Since 1 mole of NaOH reacts with 1 mole of CH3COOH, 0.06 moles of NaOH will react with 0.06 mole of CH3COOH to produce 0.06 moles of NaCH3COO and 0.06 mole of H2O.

    So, all the NaOH reacted, but 0.06 mole of CH3COOH did not react

    The final solution contains 0.06 moles of NaCH3COO, 0.06 mole of CH3COOH, and 0.06 mole of H2O.

    0.06 mole of H2O = 0.06 * 18 = 1.08 grams = 1.08 ml

    The total volume of solution = 100 ml + 200 ml + 1.08 ml = 301.08 ml = 0.30108 Liters of solution.

    3) calculate the concentration, in mol dm^-3 of the unreacted ethanoic acid in the resulting mixture?

    Remember,1 dm^3 = 1000 ml = 1 liter.

    So, 0.30108 Liters of solution. = 0.30108 dm^3

    Concentration of unreacted ethanoic acid = 0.06 mole ÷ 0.30108 dm^3 ≈ 0.2 mol dm^-3

    The concentration of NaCH3OO = 0.06 mole ÷ 0.30108 dm^3 ≈ 0.2 mol dm^-3

    4) the mixture which forms is a buffer. why does the pH of the mixture remain constant when small quantities of solutions containing H+ of OH- ions are added? ( use equations in your explanation.)

    The final solution contains equal amounts NaCH3COO and CH3COOH.

    These molecules form ions.

    NaCH3COO ↔ Na+1 + CH3COO-1

    When acid is added to the products above, the H+1 bond with CH3COO-1 ions to form CH3COOH molecules. This prevents a increase of H+1 in the solution. So the pH does not change.

    CH3COOH + H2O ↔ CH3COO-1 + H3O+1

    When base is added to the products above, the OH-1 bonds with H3O+1 to form H2O + H+. The CH3COOH-1 ion attracts the H+1 to form CH3COOH molecules. This prevents a increase of H+1 in the solution. So the pH does not change.

    These 2 process prevent the buffer solution from becoming acidic or basic.

    5) the concentration of sodium ethanoate in the mixture is 0.200 mol dm^-3. calculate the pH of the mixture, using your answer to 3

    http://chemistry.about.com/od/acidsbase1/a/henders...

    Go to the website above to see an example of determining the ph of a buffer with acetic acid.

    Henderson-Hasselbach equation for pH of a buffer

    pH = pKa + [salt] ÷ [acid]

    Ka = 1.8 * 10^-5

    pKa = -1 * log Ka = -1 * log (1.8 * 10^-5) = 4.74

    [salt] ÷ [acid] = 0.2/0.2 = 1

    pH = 4.74 + 1 = 5.74

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