Chemistry, 10.07.2019 12:30 raheeeq

Asolution of acetic acid is prepared in water by adding 11.1 g of sodium acetate to 200 ml of water. the final ph is measured to be 6.5. what are the concentrations of acetate and acetic acid in solution?

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Answer from: zazy15

Given buffer contains:

Acetic acid: CH3COOH.

Sodium Acetate : CH3COONa.

Mass of CH3COONa taken = 11.1 g

Molar mass of CH3COONa = 82.03 g/mol

# moles of CH3COONa = 11.1g/82.03 g.mol-1 = 0.1353 moles

Volume of water = 200 ml = 0.2 L

Therefore, the initial concentration of CH3COONa = 0.1353/0.2 = 0.677 M

Buffer pH = 6.5

pKa (CH3COOH) = 4.76

Based on Henderson-Hasselbalch equation:

pH = pKa + log [CH3COONa]/[CH3COOH]

Now, initially there is only 0.677 M CH3COONa . If 'x' is the concentration of the protonated species (i.e CH3COOH) formed, then the amount of CH3cooNa remaining = 0.677-x

6.5 = 4.76 + log [0.677-x]/[x]

x = 0.011 M

i.e. [CH3COOH] = 0.011 m

[CH3COONa] = 0.677-0.011 = 0.666 M

Answer from: Quest

the answer to your question is c

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³

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Asolution of acetic acid is prepared in water by adding 11.1 g of sodium acetate to 200 ml of water....

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