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mezya [45]
3 years ago
8

A 1.0L buffer solution contains 0.100 molHC2H3O2 and 0.100 molNaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5. Calculate the p

H of the solution upon the addition of 10.0 mL of 1.00 MHCl to the original buffer.
Express the your answer to two decimal places.
***will mark Brainliest for right answer ***
Chemistry
1 answer:
Crazy boy [7]3 years ago
7 0
Starting pH = pKa because the acid and conjugate base have equal concentrations.
pH = pKa = -log(1.8*10–5) = 4.74

0.0100 L • 1.00 M = 0.0100 mol HCl, which reacts with the conjugate base, producing 0.0100 mol of weak acid.

You now have 0.110 mol weak acid and 0.090 mol conjugate base. Use the Henderson-Hasselbach equation to solve for pH now.

pH = 4.74 + log(0.090/0.110) = 4.65

The equation technically says to use concentrations of the acid and base, but that doesn’t actually matter, because it would give the same answer.
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This type of bonding can be described as a "cooperation" A. lonic B. metallic C. covalent​
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Benzoic acid is a natural fungicide that naturally occurs in many fruits and berries. The sodium salt of benzoic acid, sodium be
AlexFokin [52]

Answer:

a. pH = 2.52

b. pH = 8.67

c. pH = 12.83

Explanation:

The equation of the titration between the benzoic acid and NaOH is:

C₆H₅CO₂H + OH⁻ ⇄ C₆H₅CO₂⁻ + H₂O    (1)

a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:

\eta_{NaOH} = C*V = 0.250 M*0.020 L = 5.00 \cdot 10^{-3} moles

\eta_{C_{6}H_{5}CO_{2}H}i = C*V = 0.300 M*0.050 L = 0.015 moles  

From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:

\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles

The concentration of benzoic acid is:

C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M

Now, from the dissociation equilibrium of benzoic acid we have:

C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺  

0.14 - x                            x                x

Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]}

Ka = \frac{x*x}{0.14 - x}

6.5 \cdot 10^{-5}*(0.14 - x) - x^{2} = 0   (2)  

By solving equation (2) for x we have:          

x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]

Finally, the pH is:

pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52

b. At the equivalence point, the benzoic acid has been converted to its conjugate base for the reaction with NaOH so, the equilibrium equation is:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻     (3)

The number of moles of C₆H₅CO₂⁻ is:

\eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles

The volume of NaOH added is:

V = \frac{\eta}{C} = \frac{0.015 moles}{0.250 M} = 0.060 L

The concentration of C₆H₅CO₂⁻ is:

C = \frac{\eta}{V} = \frac{0.015 moles}{(0.060 L + 0.050 L)} = 0.14 M

From the equilibrium of equation (3) we have:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻  

0.14 - x                              x               x

Kb = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}

(\frac{Kw}{Ka})*(0.14 - x) - x^{2} = 0

(\frac{1.00 \cdot 10^{-14}}{6.5 \cdot 10^{-5}})*(0.14 - x) - x^{2} = 0

By solving the equation above for x, we have:

x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]

The pH is:

pOH = -log[OH^{-}] = -log(4.64 \cdot 10^{-6}) = 5.33

pH = 14 - pOH = 14 - 5.33 = 8.67

     

c. To find the pH after the addition of 100 mL of NaOH we need to find the number of moles of NaOH:

\eta_{NaOH}i = C*V = 0.250 M*0.100 L = 0.025 moles

From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:                              

\eta_{NaOH} = \eta_{NaOH}i - \eta_{C_{6}H_{5}CO_{2}H} = 0.025 moles - 0.015 moles = 0.010 moles                          

The concentration of NaOH is:

C = \frac{\eta}{V} = \frac{0.010 moles}{0.100 L + 0.050 L} = 0.067 M

Therefore, the pH is given by this excess of NaOH:         

pOH = -log([OH^{-}]) = -log(0.067) = 1.17

pH = 14 - pOH = 12.83

I hope it helps you!    

4 0
3 years ago
Explain and illustrate the notation for distinguishing between the different p orbitals in a sublevel.
masha68 [24]

Here, the three different notation of the p-orbital in different sub-level have to generate

The value of azimuthal quantum number (l) for -p orbital is 1. We know that the magnetic quantum number m_{l} depends upon the value of l, which are -l to +l.

Thus for p-orbital the possible magnetic quantum numbers are- -1, 0, +1. So there will be three orbitals for p orbitals, which are designated as p_{x}, p_{y} and p_{z} in space.

The three p-orbital can be distinguish by the quantum numbers as-

For 2p orbitals (principal quantum number is 2)

1) n = 2, l = 1, m = -1

2) n = 2, l = 1, m = 0

3) n = 2, l = 1, m = +1

Thus the notation of different p-orbitals in the sub level are determined.  

6 0
3 years ago
Explain how the electron configuration of calcium, Ca, would illustrate the Aufbau Principle.
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Answer:

The Aufbau Principle simply helps us determine electron configuration of an atom by stating that in the ground state of an atom or ion, electrons fill subshells of the lowest available energy level, then they fill subshells of higher energy level. For example, the 1s subshell is filled before the 2s subshell is occupied. Now, when trying to figure out the electron configuration of a calcium, you need to know its atomic number to determine its amount of total electrons. Calcium has an atomic number of 20, which means it has 20 protons and 20 electrons. First remember that the "s" subshell only holds 2 electrons, the "p" subshell only hold 6 electrons, and the "d" subshell only holds up to 10 electrons. Using the Aufbau principle below, we can determine that the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s is now full we'll move to the 3p where we'll place the next six electrons. We now go to the 4s orbital where we place the remaining two electrons. With this, the calcium electron configuration will be:

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2

Hope that helps you understand!

8 0
3 years ago
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