CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
14th Edition
ISBN: 9781259137815
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 24, Problem 101AP
Interpretation Introduction

Interpretation:

The pH of the solution, obtained by dissolving the product, of the given reaction, in water, at a temperature of 25°C is to becalculated.

Concept introduction:

Equilibrium constant isdefined as the ratio of concentration of products to the concentration of reactants and is expressed as follows:

Ka=[products][reactants].

Molarity is defined as the number of moles of solute dissolved in one litre of solution and is expressed as follows:

Molarity=Moles of soluteLiter of solution.

Expert Solution & Answer
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Answer to Problem 101AP

Solution: pH is 5.72.

Explanation of Solution

Given information: The white phosphorous burnt is 10.0 g.

The volume of solution is 0.500 L.

If the white phosphorus burns in excess of oxygen, the reaction is as follows:

P4(s)+5O2(g)P4O10(s)

The product formed (P4O10) is dissolved in water and the reaction that takes place is as follows:

P4O10(s)+6H2O(l)4H3PO4(aq).

The moles formed by H3PO4 are to be calculated as follows:

Moles=Given massMolar mass.

Substitute the values in the above equation:

MH3PO4=10.0 g P4×(1 mol P4123.9 g P4)(1 mol P4O101 mol P4)(4 mol H3Po41 mol P4O10)MH3PO4=0.323 mol H3Po4.

Thus, the moles of H3PO4 are 0.323 mol  .

Molarity of the phosphoric acid solution is calculated as follows:

Molarity=Moles of soluteVolume of Solution (L).

Substitute 0.323 mol   for the moles of solute and 0.500 L for the volume of solution in the above equation as follows:

Molarity=(0.323 mol0.500 L);Molarity=0.646 M.

The equilibrium expression for a phosphoric acid solution is as follows:

                             H3PO4(aq)  +  H2O(aq)H3O+(aq)  +   H2PO(aq)initial(M):            0.646                                                   0                        0change(M):            -x                                                   +x                  +x_Equilibrium(M):  0.646 -x                                              x                        x

The value of Ka for phosphoric acid is 7.5×103, taken from the reference table 16.8.

The [H3O+] is calculated with the help of the equilibrium constant as follows:

Ka=[H3O+][H2PO4][H3PO4].

Substitute (0.646x) for the concentration of phosphoric acid, (x) for the concentration of hydronium ions, (x) for the concentration of dihydrogen phosphate ions, and 7.5×103 for Ka in the above equation as follows:

7.5×103=(x)(x)(0.646x)7.5×103=(x2)(0.646x)x2=(7.5×103)(0.646x).

Rearrange the above equation:

(x2+7.5×103x4.85×103)=0.

This is a quadratic equation, by solving:

x=0.066 M.Therefore, [H3O+]=0.066 M.

The pH of the solution is to be calculated as follows:

pH=log[H+].

Here, the concentration of hydronium ions ([H3O+]) is equal to the concentration of [H+].

pH =log(0.066)pH=1.18.

Conclusion

The pH of the solution at 25°C is 1.18.

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Chapter 24 Solutions

CHEMISTRY >CUSTOM<

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