Chemistry, 27.07.2019 13:00 datands

Abuffer is prepared by adding 20.0 g of sodium acetate 1ch3coona2 to 500 ml of a 0.150 m acetic acid 1ch3cooh2 solution. (a) determine the ph of the buffer. (b) write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the

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Answer from: k11kiki

First, we have to get moles of CH3COONa = mass/molar mass

= 20 g / 82.03 g/mol = 0.244 moles

when we have [CH3COOH] = 0.15 M

∴ [CH3COONa] = moles of CH3COONa / Volume of solution

= 0.244 moles / 0.5L = 0.488 M

when we look up for Ka of acetic acid value it is equal 1.8 x 10^-5

So we can get Pka = -㏒Ka

= -㏒(1.8 x10^-5)
= 4.7

now we will use Henderson - Hasselbalchn equation to get the PH:

PH = Pka + ㏒[conjugate basic/weak acid]

when CH3COOH is the weak acid & CH3COO- is the conjugate base so by substitution:

PH = 4.7 + ㏒ (0.488/0.15)

= 5.2

b) when we have this equation for the reaction:

HCl + CH3COONa → CH3COOH + NaCl

ionic equation : H+ + Cl- + CH3COO- + Na+ → CH3COOH + Na+ + Cl-

when HCl + H2O → H3O+ + Cl-

∴ the reaction will be:

CH3COO- (aq) + H3O+(aq) → CH3COOH(aq) + H2O(l)

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

Answer from: Quest

It will appear as o2(g) as a reactant

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Abuffer is prepared by adding 20.0 g of sodium acetate 1ch3coona2 to 500 ml of a 0.150 m acetic acid...

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