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nevsk [136]
3 years ago
11

Calculate the pOH and the pH of a 5.0 x 10-2 M solution of NaOH.

Chemistry
1 answer:
k0ka [10]3 years ago
5 0

Answer:

pOH = 1.3, pH = 12.7

Explanation:

Since NaOH is a strong base, it will completely ionize; further, since it completely ionizes, our hydroxide concentration (a product of the ionization) will be the same as the given concentration of NaOH.

NaOH -> Na⁺ + OH⁻, [OH⁻] = 5.0 x 10^-2 M

pOH is the negative log of the hydroxide concentration, so plug our hydroxide concentration in:

pOH = -log[OH⁻] = -log[5.0 x 10^-2 M] = 1.3

Since pH + pOH = 14, we can plug in pOH and solve for pH:

pH + 1.3 = 14

pH = 14 - 1.3 = 12.7

Thus, our pOH = 1.3 and pH = 12.7.

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ahrayia [7]
<span>Answer: 100 ml
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<span>Explanation:


1) Convert 1.38 g of Fe₂S₃ into number of moles, n


</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
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iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
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<span>2) Use the percent yield to calculate the theoretical amount:
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<span>65% = 0.65 = actual yield/ theoretical yield =>


</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>

3) Chemical equation:
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<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)


4) Stoichiometrical mole ratios:
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<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl


5) Proportionality:


</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
<span>
=> x = 0.020 mol FeCl₃


6) convert 0.020 mol to volume
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<span>i) Molarity formula: M = n / V
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<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
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At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:
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<u>Answer:</u> The equilibrium concentration of ICl is 0.27 M

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.45 moles

Initial moles of chlorine gas = 0.45 moles

Volume of the flask = 2.0 L

The molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}

Initial concentration of iodine gas = \frac{0.45}{2}=0.225M

Initial concentration of chlorine gas = \frac{0.45}{2}=0.225M

For the given chemical equation:

2ICl(g)\rightarrow I_2(g)+Cl_2(g);K_c=0.11

As, the initial moles of iodine and chlorine are given. So, the reaction will proceed backwards.

The chemical equation becomes:

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<u>At eqllm:</u>   0.225-x    0.225-x     2x

The expression of K_c for above equation follows:

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Putting values in above equation, we get:

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Neglecting the value of x = 0.668 because equilibrium concentration cannot be greater than the initial concentration

So, equilibrium concentration of ICl = 2x = (2 × 0.135) = 0.27 M

Hence, the equilibrium concentration of ICl is 0.27 M

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