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nekit [7.7K]
3 years ago
12

Acetic acid is a weak acid with a pKa of 4.76. What is the concentration of acetate in a buffer solution of 0.2M at pH 4.9. Give

your answer in M but as a numeral only to 2 decimal places.
Chemistry
1 answer:
Ghella [55]3 years ago
6 0

Answer:

[base]=0.28M

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:

pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\

[base]=1.38[acid]=1.38*0.20M=0.28M

Regards.

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Under what conditions will deviations from the "ideal" gas be expected?
mixas84 [53]

Answer:

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1. Low Temperature

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3 0
3 years ago
Cho m gam FeO tác dụng hết với dung dịch H2SO4, thu được 200 ml dung dịch FeSO4 1M. Giá trị của m là
BaLLatris [955]

<u>Answer:</u> The mass of FeO required is 14.37 g

<u>Explanation:</u>

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume}} ......(1)

We are given:  

Molarity of iron (II) sulfate = 1 M

Volume of solution = 200 mL = 0.200 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

\text{Moles of }FeSO_4=(1mol/L\times 0.200L)=0.200mol

The chemical equation for the reaction of FeO with sulfuric acid follows:

FeO+H_2SO_4\rightarrow FeSO_4+H_2O

By stoichiometry of the reaction:

If 1 mole of iron (II) sulfate is produced by 1 mole of FeO

So, 0.200 moles of iron (II) sulfate will produce = \frac{1}{1}\times 0.200=0.200mol of FeO

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We know, molar mass of FeO = 71.84 g/mol

Putting values in above equation, we get:

\text{Mass of }FeO=(0.200mol\times 71.84g/mol)=14.37g

Hence, the mass of FeO required is 14.37 g

4 0
3 years ago
What is the maximum mass, in kilograms, of (NH4)2U2O7 that can be formed from the reaction of 100 kg of water and 100 kg of ammo
V125BC [204]

Mass of (NH₄)₂U₂O₇ : 410.05 kg

<h3>Further explanation</h3>

Reaction

2UO₂SO₄ + 6NH₃ + 3H₂O → (NH₄)₂U₂O₇ + 2(NH₄)₂SO₄

MW UO₂SO₄ :  366.091

MW (NH₄)₂U₂O₇ : 624.131

MW H₂O :  18.0153

MW NH₃ : 17.0306

mol of 100 kg water :

\tt \dfrac{100}{18.0153}=5.55

mol of 100 kg ammonia :

\tt \dfrac{100}{17.036}=5.87

mol of UO₂SO₄ :

\tt \dfrac{481}{366.091}=1.314

Limiting reactants : smallest mol ratio(mol : coefficient)

\tt \dfrac{5.55}{3}\div \dfrac{5.87}{6}\div \dfrac{1.314}{2}=1.85\div 0.98\div 0.657

UO₂SO₄ ⇒ Limiting reactants

mol (NH₄)₂U₂O₇ : mol UO₂SO₄

\tt \dfrac{1}{2}\times 1.314=0.657

mass (NH₄)₂U₂O₇

\tt 0.657\times 624.131=410.05

5 0
3 years ago
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