Chemistry, 23.11.2020 21:10 Vimmia

The average titration volume of the 0.09876 M NaOH used in back titration is 29.59 ml. Calculate the number of mmoles of HCl in the 250 ml volumetric flask.

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Answer from: nyceastcoast

The number of moles of HCl in the 250 mL volumetric flask is 0.003 moles

Explanation:

Firstly, we solve for the concentration of acid using the formula

CaVa/CbVb = nₐ/nb

where Ca is the concentration of acid

Cb is the concentration of base

Va is the volume of acid

Vb is the volume of base

nₐ is the number of moles of acid (from the equation)

nb is the number of moles of base (from the equation)

Ca × 250/0.09876 × 29.59 = 1/1

Ca = 0.09876 × 29.59/250

Ca = 0.012 M

To determine the number of moles of HCl acid present in the 250 ml volumetric flask, the formula for molarity is used

Molarity = number of moles ÷ volume (in liter or dm³)

Volume needs to be converted to liter; 250 ml ⇒ 0.25 L

Molarity of the acid is 0.012 M

From the formula above, number of moles = molarity × volume (in liter)

number of moles = 0.012 × 0.25

number of moles of acid = 0.003 moles

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

Answer(28.02g/mol);

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The average titration volume of the 0.09876 M NaOH used in back titration is 29.59 ml. Calculate th...

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