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sergejj [24]
3 years ago
14

When a nonmetal bonds with a nonmetal it is a _ bond

Chemistry
1 answer:
Karolina [17]3 years ago
3 0

It is a covalent bond

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6. A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCI. Calculate the pH
Vadim26 [7]

Answer:

a) pH = 9.14

b) pH = 8.98

c) pH = 8.79

Explanation:

In this case we have an acid base titration. We have a weak base in this case the pyridine (C₅H₅N) and a strong acid which is the HCl.

Now, we want the know the pH of the resulting solution when we add the following volume of acid: 0, 10 and 20.

To know this, we first need to know the equivalence point of this titration. This can be known using the following expression:

M₁V₁ = M₂V₂  (1)

Using this expression, we can calculate the volume of acid required to reach the equivalence point. Doing that we have:

M₁V₁ = M₂V₂

V₁ = M₂V₂ / M₁

V₁ = 0.125 * 25 / 0.1 = 31.25 mL

This means that the acid and base will reach the equivalence point at 31.25 mL of acid added. So, the volume of added acid of before, are all below this mark, so we can expect that the pH of this solution will be higher than 7, in other words, still basic.

To know the value of pH, we need to apply the following expression:

pH = 14 - pOH  (2)

the pOH can be calculated using this expression:

pOH = -log[OH⁻]  (3)

The [OH⁻] is a value that can be calculated when the pyridine is dissociated into it's ion. However, as this is a weak acid, the pyridine will not dissociate completely in solution, instead, only a part of it will be dissociated. Now, to know this, we need the Kb value of the pyridine.

The reported Kb value of the pyridine is 1.5x10⁻⁹ so, with this value we will do an ICE chart for each case, and then, calculate the value of the pH.

<u>a) 0 mL of acid added.</u>

In this case, the titration has not begun, so the concentration of the base will not be altered. Now, with the Kb value, let's write an ICE chart to calculate the [OH⁻], the pOH and then the pH:

       C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.125                                0             0

e)        -x                                   +x           +x

c)      0.125-x                              x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.125-x --> Kb is really small, so we can assume that x will be very small too, and 0.125-x can be neglected to only 0.125, and then:

1.5x10⁻⁹ = x² / 0.125

1.5x10⁻⁹ * 0.125 = x²

x = [OH⁻] = 1.37x10⁻⁵ M

Now, we can calculate the pOH:

pOH = -log(1.37x10⁻⁵) = 4.86

Finally the pH:

pH = 14 - 4.86

<h2>pH = 9.14</h2>

<u>b) 10 mL of acid added</u>

In this case the titration has begun so the acid starts to react with the base, so we need to know how many moles of the base remains after the volume of added acid:

moles acid = 0.1 * (0.010) = 1x10⁻³ moles

moles base = 0.125 * 0.025 = 3.125x10⁻³

This means that the base is still in higher quantities, and the acid is the limiting reactant here, so the remaining moles will be:

remaining moles of pyridine = 3.125x10⁻³ - 1x10⁻³ = 2.125x10⁻³ moles

The concentration of pyridine in solution:

[C₅H₅N] = 2.125x10⁻³ / (0.025 + 0.010) = 0.0607 M

Now with this concentration, we will do the same procedure of before, with the ICE chart, but replacing this new value of the base, to get the [OH⁻] and then the pH:

        C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.0607                             0             0

e)        -x                                   +x           +x

c)      0.0607-x                           x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.0607-x --> 0.0607

1.5x10⁻⁹ = x² / 0.0607

1.5x10⁻⁹ * 0.0607 = x²

x = [OH⁻] = 9.54x10⁻⁶ M

Now, we can calculate the pOH:

pOH = -log(9.54x10⁻⁶) = 5.02

Finally the pH:

pH = 14 - 5.02

<h2>pH = 8.98</h2>

<u>c) 20 mL of acid added:</u>

In this case the titration it's almost reaching the equivalence point and the acid is still reacting with the base, so we need to know how many moles of the base remains after the volume of added acid:

moles acid = 0.1 * (0.020) = 2x10⁻³ moles

moles base = 0.125 * 0.025 = 3.125x10⁻³

This means that the base is still in higher quantities, and the acid is the limiting reactant here, so the remaining moles will be:

remaining moles of pyridine = 3.125x10⁻³ - 2x10⁻³ = 1.125x10⁻³ moles

The concentration of pyridine in solution:

[C₅H₅N] = 1.125x10⁻³ / (0.025 + 0.020) = 0.025 M

Now with this concentration, we will do the same procedure of before, with the ICE chart, but replacing this new value of the base, to get the [OH⁻] and then the pH:

        C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.025                                0             0

e)        -x                                   +x           +x

c)      0.025-x                             x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.025-x --> 0.025

1.5x10⁻⁹ = x² / 0.025

1.5x10⁻⁹ * 0.025 = x²

x = [OH⁻] = 6.12x10⁻⁶ M

Now, we can calculate the pOH:

pOH = -log(6.12x10⁻⁶) = 5.21

Finally the pH:

pH = 14 - 5.21

<h2>pH = 8.79</h2>
5 0
3 years ago
Question 3 of 11
daser333 [38]

Answer:

c

Explanation:

b and d are out, the variables are changed. a would be a repetition, not a replication. c uses the same method and variables with a different control group

6 0
3 years ago
How many milliners of hydrogen gas ar produced by the reaction 0.020 moles of magnesium with excess of hydrochloride acid at sto
jeyben [28]

Answer:- 448 mL of hydrogen gas are formed.

Solution:- It asks to calculate the volume of hydrogen gas formed in milliliters at STP when 0.020 moles of magnesium reacts with excess HCl acid. The balanced equation is:

Mg+2HCl\rightarrow MgCl_2+H_2(g)

There is 1:1 mol ratio between Mg and hydrogen gas. So, the moles of hydrogen gas is also equals to the moles of Mg reacted.

moles of Hydrogen gas formed = 0.020 mol

At STP, volume of 1 mol of the gas is 22.4 L. We need to calculate the volume of 0.02 moles of hydrogen gas.

0.02mol(\frac{22.4L}{1mol})

= 0.448 L

They want answer in mL. So, let's convert L to mL using the conversion formula, 1L = 1000mL

0.448L(\frac{1000mL}{1L})

= 448 mL

So, 0.020 moles of magnesium would produce 448 mL of hydrogen gas at STP on reacting with excess of HCl acid.

5 0
3 years ago
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