Chemistry, 04.07.2019 00:00 Violinartist7267

Calculate the ph during the titration of 10.00 ml of 0.400 m hypochlorous acid with 0.500 m naoh. first what is the initial ph (before any naoh is added)? the ka for hocl is 3.0 x 10-8 m.

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Answer from: GamerGirl15

3.96

Explanation :

Set up an ICE table

HClO + H₂O ⇌ H₃O⁺ + ClO⁻

I: 0.400 0 0

C: -x +x +x

E: 0.400-x x x

Kₐ = {[H₃O⁺][ClO⁻]}/[HClO] = 3.0 × 10⁻⁸ Substitute values

(x × x)/(0.400-x) = 3.0 × 10⁻⁸ Combine like terms

x²/(0.400-x) Check that x ≪ 0.400

0.400/(3.0 × 10⁻⁸) = 1.3 × 10⁷ > 400 x ≪ 0.400; Ignore x

x²/0.400 = 3.0 × 10⁻⁸ Cross-multiply

x² = 0.400 × 3.0 × 10⁻⁸ Complete the multiplication

x² = 1.20 × 10⁻⁸ Take the square root of both sides

x = [H₃O⁺] = 1.10 × 10⁻⁴ Take the negative log of each side

-log[H₃O⁺] = pH = -log(1.10 × 10⁻⁴) Complete the operation

pH = 3.96

Answer from: Quest

answer is: c. the experiment shows that the red substance experienced a chemical change.

this is chemical change (chemical reaction), because two new substances are formed (metalig liquid and gas), the atoms are rearranged and the reaction is followed by an energy change.

this is example of chemical decomposition.

chemical decomposition (analysis) is the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.

chemical decomposition is opposite of chemical synthesis.

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.

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Calculate the ph during the titration of 10.00 ml of 0.400 m hypochlorous acid with 0.500 m naoh. fi...

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