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3 years ago

Calculate either [H3O ] or [OH–] for the solutions below at 25 C.

2 answers:

5 0

Answer is: concentration of hydronium ions is 5,5·10⁻⁸ M.
[OH⁻] = 1,83·10⁻⁷ M; equilibrium concentration.
The Kw (the ionic product for water) at 25°C is 1·10⁻¹⁴ mol²/dm⁶ or 1·10⁻¹⁴ M².<span>
Kw = [H</span>₃O⁺] · [OH⁻].
[H₃O⁺] = Kw ÷ [OH⁻].
[H₃O⁺] = 1·10⁻¹⁴ M² ÷ 1,83·10⁻⁷ M.
[H₃O⁺] = 5,5·10⁻⁸ M.

Fittoniya [83]3 years ago

5 0

This problem requires using the concept of pH and pOH. The “p” stands for the negative log so pH and pOH represent the negative log of the concentration of hydrogen or hydroxide ions.

Here is the solution:

pOH = -log [OH-]

pOH = -log [1.83x10^-7 M]

pOH = 6.74

pH + pOH = 14

pH = 14 - 6.74

pH = 7.26

pH = -log [H3O+]

7.26 = -log[H3O+]

[H3O+] = 5.46 x 10^-8 M

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3 years ago

I need help with this plz ???

saw5 [17]

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3 years ago

Problem page gaseous ethane ch3ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o

iVinArrow [24]

$m(\text{CO}_2) = 2.24 \; \text{g}$

Ethene react with oxygen at a $2 : 7$ molar ratio:

$2\; \text{C}_2 \text{H}_6 (g) + 7\; \text{O}_2 (g) \to 6\; \text{H}_2{O} (g) + 4\; \text{CO}_2 (g)$

Convert the quantity of each reactant supplied to number of moles of particles:

$n(\text{C}_2\text{H}_6) = 0.60 \; \text{g} / 28.05 \; \text{g} \cdot \text{mol}^{-1} = 0.0214 \; \text{mol}$
$n(\text{O}_2) = 3.27 \; \text{g} / 32.00 \; \text{g} \cdot \text{mol}^{-1} = 0.102 \; \text{mol}$

The question stated not whether both reactants were used up in this process. Thus start by testing the assumption that e.g., ethene was used up while some oxygen gas were left unreacted (ethene as the <em>limiting </em>reagent.) Under this assumption, the relative availability of the two species, $n(\text{C}_2 \text{H}_6) /2$ and $n(\text{O}_2) /7$ (as seen in the balanced chemical equation) shall satisfy the relationship

$n(\text{O}_2) / 7 - n(\text{C}_2 \text{H}_6) / 2 > 0$

In other words,

$n(\text{O}_2)/7 > n(\text{C}_2 \text{H}_6)/2$

$n(\text{C}_2 \text{H}_6) / n(\text{O}_2) < 2/7 \approx 0.286$

Evaluating the expression $n(\text{C}_2 \text{H}_6) / n(\text{O}_2)$ with data given in the question yields approximately $0.210 < 0.286$ , which does satisfy the relationship. Hence the assumption holds and ethene is the limiting reactant.

The quantity of a reactant produced in a chemical reaction is related to its stoichiometric (of relating to proportions) relationship with the limiting reactant (or any of the reactants in case of more than one limiting reactant.) For this scenario, given the molar ratio $n(\text{C}_2\text{H}_6) : n( \text{CO}_2) = 2:4$ ,