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Anna007 [38]
3 years ago
6

Find the pH of a 0.010 M HNO2 solution.

Chemistry
1 answer:
lidiya [134]3 years ago
6 0
Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol

M (molarity) = 0.010 M (Mol/L)


Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) = 5.0*10^{-4}
\alpha^2 (degree\:of\:ionization) = ?

Ka = M * \alpha^2
5.0*10^{-4} = 0.010* \alpha^2
0.010\alpha^2 = 5.0*10^{-4}
\alpha^2 = \frac{5.0*10^{-4}}{0.010}
\alpha^2\approx500*10^{-4}

\alpha\approx\sqrt{500*10^{-4}}
\alpha \approx 2.23*10^{-3}

Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:

[ H_{3} O^+] = M* \alpha
[ H_{3} O^+] = 0.010* 2.23*10^{-3}
[ H_{3} O^+] \approx 0.0223*10^{-3}
[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:
log10(2.23) ≈ 0.34
pH = ?
[ H_{3} O^+] = 2.23*10^{-5}

Formula:
pH = - log[H_{3} O^+]

Solving:
pH = - log[H_{3} O^+]
pH = -log2.23*10^{-5}
pH = 5 - log2.23
pH = 5 - 0.34
\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark

Note:. The pH <7, then we have an acidic solution.
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Balancing Chemical Equations
Citrus2011 [14]

Start by assigning the most complicated species a coefficient of one. That species should contain the greatest number of elements, e.g. \text{HCl}.

\text{Zn} + \text{HCl} \to 1\; \text{ZnCl}_2 + \text{H}_2 <em>(not balanced)</em>

Assign coefficients to the rest of the species based on the conservation of atoms. For instance, the left hand side of the equation now contains one atom of hydrogen H and one atoms of chlorine Cl. The left hand side shall have an identical configuration. Both zinc chloride \text{ZnCl}_2 and hydrogen \text{H}_2 should therefore have a coefficient of 1/2. (Don't panic about the fractions. They are to be eliminated in a few more steps.)

\text{Zn} + 1\; \text{HCl} \to 1/2 \; \text{ZnCl}_2 + 1/2 \; \text{H}_2 <em>(not balanced)</em>

The coefficient 1/2 in front of zinc chloride indicates the presence of 1/2 zinc atom in the right hand side of the equation. Zinc on the left hand side of the equation should accordingly have a coefficient of 1/2.

1/2 \; \text{Zn} + 1\; \text{HCl} \to 1/2 \; \text{ZnCl}_2 + 1/2 \; \text{H}_2

Increasing coefficients on both sides of the equation by a factor of two to eliminate all fractions. Hence the balanced equation.

1 \; \text{Zn} + 2 \; \text{HCl} \to 1 \; \text{ZnCl}_2 + 1 \; \text{H}_2 (balanced)

The same set of operations should work for the second equation.

4 \; \text{Fe} + 3\; \text{O}_2 \to 2\; \text{Fe}_2\text{O}_3 (balanced)

Note that the third equation does not accurately represent the catalytic decomposition of hydrogen peroxide \text{H}_2\text{O}_2. The balanced equation should be:

2\; \text{H}_2\text{O}_2 \to 2\; \text{H}_2\text{O} + \text{O}_2 (balanced)


6 0
3 years ago
A 25.00-ml sample of propionic acid, hc3h5o2, of unknown concentration was titrated with 0.141 m koh. the equivalence point was
Scorpion4ik [409]
 The concentration of propionate ions   at  equivalent point is  0. 084 M
  calculation
write the equation for reaction
HC3H5O2   +  KOH  → H2O  + KC3H5O2

find the moles  of  KOH  used
moles=  molarity  × volume in liters
volume  in liters = 37.48 ml/1000 =0.03718L
molarity = 0.141 M


moles is therefore =  0.141 M  x0.03718 =  5.24238  x10^-3 moles

by use of mole  ratio between  KOH  and K3C3H5O2  which is 1:1  the  moles of K3C3H5O2  is also = 5.24238 x10^-3  moles

molarity is therefore = number  of moles/volume in liters
volume  in liters = 25/1000 +37.48/1000=0.06248 L

molarity of  propionate ions is therefore = (5.24238 x10^-3) / 0.06248 = 0.084  M
7 0
2 years ago
The songs of birds are part of a ritual to attractpotential partners of their own species. The song presents specific patterns r
liberstina [14]

Answer:

Explanation:

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6 0
3 years ago
what would be the effect on the properties of the water molecule if oxygen and hydrogen had equal electronegativity?
ohaa [14]

Answer:

It would no longer be a polar molecule with a slightly positive and slightly negative ends. It then would not be able to dissolve ionic salts or polarized organic compounds such as sugar.

Explanation:

7 0
2 years ago
6.0 mol NaOH reacts with
lina2011 [118]

Taking into account the reaction stoichiometry, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 NaOH + H₃PO₄ → 3 H₂O + Na₃PO₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • NaOH: 3 moles
  • H₃PO₄: 1 mole
  • H₂O: 3 moles
  • Na₃PO₄: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of H₃PO₄ reacts with 3 moles of NaOH, 9 moles of H₃PO₄ reacts with how many moles of NaOH?

moles of NaOH=\frac{9 moles of H_{3} PO_{4} x3 moles of NaOH}{1 mole of H_{3} PO_{4}}

moles of NaOH= 27 moles

But 27 moles of NaOH are not available, 6 moles are available. Since you have less moles than you need to react with 9 moles of H₃PO₄, NaOH will be the limiting reagent.

<h3>Moles of Na₃PO₄ formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of NaOH form 1 mole of Na₃PO₄, 6 moles of NaOH form how many moles of Na₃PO₄?

moles of Na_{3}P O_{4} =\frac{6  moles of NaOHx1 mole of Na_{3}P O_{4} }{3 moles of NaOH}

<u><em>moles of Na₃PO₄= 2 moles</em></u>

Then, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

8 0
2 years ago
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