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# Tutor profile: Ian B.

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Ian B.
Physical/Quantum Chemistry undergraduate student, tutor for five years
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## Questions

### Subject:Spanish

TutorMe
Question:

Translate the following sentence into Spanish and name the grammatical construction used. If Antonio had not adopted (adoptar) the dog, he would not have met his best friend.

Inactive
Ian B.

This is a past contrary-to-fact construction: Si Antonio no hubiera adoptado a su perro, (él) no habría conocido a su mejor amigo.

### Subject:Physical Chemistry

TutorMe
Question:

Derive an expression for the average distance from the nucleus of an electron in the 3p orbital. You may leave your answer in the form of an integral.

Inactive
Ian B.

The average radius $$<r>$$ can be determined from the expectation value of the $$r$$ operator on the orbital wave function: $$<r> = \frac{\int_{all space}\Psi^*r\Psi d\tau}{\int_{all space} \Psi^*\Psi d\tau}$$ Since the orbital wave function is normalized, this simplifies to: $$<r> = \int_{all space}\Psi^*r\Psi d\tau$$ The volume element can be written in spherical coordinates as: $$<r> = \int_0^\infty\int_0^\pi\int_0^{2\pi}\Psi^*r\Psi r^2dr\sin(\theta) d\theta d\phi$$ Orbital wave functions are products of the hydrogenic radial wave functions with a dependence on $$r$$ and the spherical harmonic wave functions with a dependence on $$\theta$$ and $$\phi$$, allowing our expression to be rewritten as: $$<r> = \int_0^\infty\int_0^\pi\int_0^{2\pi}r^3R^2(r)dr\Upsilon^*(\theta,\phi)\Upsilon(\theta,\phi)\sin(\theta) d\theta d\phi$$ The spherical harmonic wave functions are normalized, simplifying our expression to: $$<r> = \int_0^\infty\int_0^\pi\int_0^{2\pi}r^3R^2(r)dr$$ The hydrogenic radial wave function of the 3p orbital is equal to: $$\frac{1}{27(6)^{1/2}}(\frac{Z}{\alpha_o})(4-\frac{2Zr}{3\alpha_o})(\frac{2Zr}{\alpha_o})e^{\frac{Zr}{3\alpha_o}}$$ This can now be inserted into the integral, yielding: $$<r> = \int_o^\infty r^3(\frac{1}{27(6)^{1/2}}(\frac{Z}{\alpha_o})(4-\frac{2Zr}{3\alpha_o})(\frac{2Zr}{\alpha_o})e^{\frac{Zr}{3\alpha_o}})^2 dr$$ The answer to this integral would yield the average value of an electron in the 3p orbital from the nucleus.

### Subject:Chemistry

TutorMe
Question:

Derive an expression for the determination of the pH of a solution of HA of molarity M and volume V upon the addition of m moles of NaA. Assume that the volume stays relatively constant. You may use the $$K_a$$ of the acid in your answer.

Inactive
Ian B.

The equilibrium is written as: $$HA (aq) + H_2O (l) \rightleftharpoons H_3O^+ (aq) + A^- (aq)$$ The $$K_a$$ of a reaction, disregarding activity, is equal to the concentrations of each product raised to its stochiometric coefficient divided by the same values for the reactants. Therefore, the expression for the $$K_a$$ of this acid is: $$K_a = \frac{[H_3O^+][A^-]}{[HA]}$$ The pH of a solution is defined as: $$pH = -log[H^+] = -log[H_3O^+]$$ Therefore, we need to rearrange our $$K_a$$ equation for the concentration of the hydronium ion: $$[H_3O^+] = \frac{K_a[HA]}{[A^-]}$$ Plugging this into our equation for the pH yields: $$pH = -log(\frac{K_a[HA]}{[A^-]})$$ To finish our derivation, we need to write everything in this expression in terms of everything given in the problem. The concentration of the acid HA is already given as M. Since NaA fully dissociates in solution due to the cation being sodium, its concentration is equal to the concentration of NaA, which can be calculated by simply dividing by the moles by the volume: $$[A^-] = [NaA] = \frac{m}{V}$$ Our final expression, therefore, is: $$pH = -log(\frac{K_a[M]}{[\frac{m}{V}]})$$

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