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stellarik [79]
3 years ago
7

Calculate the pZn of a solution prepared by mixing 25.0 mL of 0.0100 M EDTA with 50.0 mL of 0.00500 M Zn2 . Assume that both the

Zn2 and EDTA solutions are buffered with 0.100 M NH3 and 0.176 M NH4Cl.
Chemistry
1 answer:
egoroff_w [7]3 years ago
4 0

Answer:

\mathbf{pZn ^{2+} =8.8569 }

Explanation:

Using the approach of Henderson-HasselBalch equation, we have :

pH = pKa[NH^+_4] + log \dfrac{[NH_3]}{[NH_4^+]}

where;

the pKa of NH^+_4 = 9.26

concentration of NH_3 = 0.100 M

concentration of NH_4Cl = 0.176 M

∴

the pH of the buffered solution is :

pH = 9.26 + log \dfrac{[0.100]}{[0.176]}

pH = 9.26 + log (0.5682)

pH = 9.26 +(-0.2455)

pH =9.02

The Chemical equation for the reaction of Zn ^{2+} and EDTA is :

Zn^{2+}_{(aq)} + Y^{4-}_{(aq)}  \iff ZnY^{2-} _{(aq)}

Here;

Y^{4-}_{(aq)} denotes the fully deprotonated form of the EDTA

The formation constant K_f of the equation for the reaction can be represented as:

K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ][Y^{4-}]}      ----- (1)

The logarithm of the formation constant of Zn - EDTA complex = 16.5

K_f  = 10^{16.5}

K_f  = 3.16  \times 10^{16}

Since the formation constant in the above equation signifies that the EDTA is present in  Y^{4-},

Then:

\alpha _{Y^{4-} }= \dfrac{Y^{4-}}{C_{EDTA}}

{Y^{4-}}= \alpha_ {Y^{4-}} \times {C_{EDTA}}

From (1)

K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ][Y^{4-}]}  

K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ \  \alpha_ {Y^{4-}} \times {C_{EDTA}}}

∴

K_f' = K_f \times \alpha _Y{^4-} = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }

where;

K_f' = conditional formation constant

\alpha _Y{^4-} = the fraction of EDTA that exit in the form of the presences of the 4 charges .

So at equivalence point :

all the Zn^{2+} initially in titrand  is now present in ZnY^{2-}

K_f' = K_f \times \alpha _Y{^4-}

Obtaining the data for the value of \alpha _Y{^4-} at the reference table:

\alpha _Y{^4-}  =  5.4 \times 10^{-12}

∴

K_f' =  3.16 \times 10^{16} \times 5.4 \times 10^{-2}

K_f' =  1.7064 \times 10^{15}

To calculate the moles of  EDTA ,Zn^{2+}  , ZnY^{2-} ; we have:

moles of  EDTA = 0.0100 M × 0.025 L

moles of  EDTA = 2.5 \times 10^{-4} \ mole

moles of Zn^{2+} = 0.00500 M  × 0.050 L

moles of Zn^{2+} = 2.5 \times 10^{-4} \ mole

moles of  ZnY^{2-}  =  \dfrac{initial \ mole}{total \ volume}

moles of  ZnY^{2-}  = \dfrac{2.5 \times 10^{-4}}{ 0.025 + 0.050 }

moles of  ZnY^{2-}  = \dfrac{2.5 \times 10^{-4}}{ 0.075 }

moles of  ZnY^{2-}  = 0.0033333 M

Recall that:

K_f' = K_f \times \alpha _Y{^4-} = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }

K_f' = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }

Assume Q² is the amount of complex dissociated in ZnY^{2-}

ZnY^{2-}  \iff Zn^{2+} + C_{EDTA}  

i.e Q^2 = Zn^{2+} + C_{EDTA}

1.707 \times 10^{15}= \dfrac{0.0033333}{Q}

Q= \dfrac{0.0033333}{1.707 \times 10^{15}}

Q^2= \dfrac{0.0033333}{1.707 \times 10^{15}}

Q^2= 1.9527 \times 10^{-18}

Q= \sqrt{1.9527 \times 10^{-18}}

Q = 1.397 \times 10^{-9} M

[Zn^{2+}]= 1.39 \times 10^{-9} \ M

∴

pZn ^{2+} =- log [Zn^{2+}]

pZn ^{2+} = -  log (1.39 \times 10^{-9} ) \ M

\mathbf{pZn ^{2+} =8.8569 }

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Dvinal [7]
Since the question manages to include moles, pressure, volume, and temperature, then it is evident that in order to find the answer we will have to use the Ideal Gas Equation:  PV = nRT (where P = pressure; V = volume; n = number of moles; R = the Universal Constant [0.082 L·atm/mol·K]; and temperature.

First, in order to work out the questions, there is a need to convert the volume to Litres and the temperature to Kelvin based on the equation:
         250 mL = 0.250 L
             58 °C = 331 K

Also, based on the equation   P = nRT ÷ V

⇒         P  = (2.48 mol)(0.082 L · atm/mol · K)(331 K)  ÷  0.250 L
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7 0
3 years ago
3Al + 3NH4ClO4 → Al2O3 + AlCl3 + 3NO + 6H2O<br> How many moles of AlCl3 are produced?
zhuklara [117]

Answer: 1 mol of AlCl_3 will be produced from this reaction.

Explanation: Reaction follows,

3Al+3NH_4ClO_4\rightarrow Al_2O_3+AlCl_3+3NO+6H_2Oc

As seen from the balanced chemical equation above, we get

For every 3 moles of Aluminium and 3 moles of NH_4ClO_4, 1 mole of Al_2O_3 is formed.

For every 3 moles of Aluminium and 3 moles of NH_4ClO_4, 1 mole of AlCl_3 is formed.

For every 3 moles of Aluminium and 3 moles of NH_4ClO_4, 3 moles of NO is formed.

For every 3 moles of Aluminium and 3 moles of NH_4ClO_4, 6 moles of H_2O is formed.

7 0
3 years ago
Read 3 more answers
What is the percent composition of NaHCO3?
zhuklara [117]

Answer:

                Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)

Explanation:

<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.

Calculating Percent Composition of NaHCO₃:

1: Calculating Molar Masses of all elements present in NaHCO₃:

              a) Na  =  22.99 g/mol

             b) H  =  1.01 g/mol

              c) C  =  12.01 g/mol

              d) O₃  =  16.0 × 3 =  48 g/mol

2: Calculating Molecular Mass of NaHCO₃:

              Na  =  22.99 g/mol

             H    =  1.01 g/mol

              C    =  12.01 g/mol

              O₃  =  48 g/mol

                       ----------------------------------  

Total                  84.01 g/mol

3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:

For Na:

                 =  22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  27.36 %

For H:

                 =  1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  1.20 %

For C:

                 =  12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  14.29 % ≈ 14.30 %

For O:

                 =  48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  57.13 % ≈ 57.14 %

8 0
3 years ago
A student ran the following reaction in the laboratory at 671 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 7.33×10-2 moles of N
vaieri [72.5K]

Answer:

Kc = 8.05x10⁻³

Explanation:

This is the equilibrium:

                 2NH₃(g)   ⇄     N₂(g)     +     3H₂(g)

Initially       0.0733

React         0.0733α          α/2                3/2α

Eq     0.0733 - 0.0733α    α/2                0.103

We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.

Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.

3/2α = 0.103

α = 0.103 . 2/3 ⇒ 0.0686

So, concentration in equilibrium are

NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682

N₂ = 0.0686/2 = 0.0343

So this moles, are in a volume of 1L, so they are molar concentrations.

Let's make Kc expression:

Kc= [N₂] . [H₂]³ / [NH₃]²

Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³

3 0
3 years ago
The mass of cupric sulfate needed to make .5 liters of a .5M solution is
dalvyx [7]

Answer:

C

Explanation:

3 0
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