Chemistry 2720 Fall 2000 Assignment 1 Solutions

Nomenclature

1.

Na₂HPO₄

2.

XeF₄

3.

Fe₂O₃

4.

Co²⁺

5.

HCl

6.

N₂O₅

7.

F₂

8.

CaBr₂

9.

O₂

10.

NH₃

11.

I^-

12.

NH₄NO₃

13.

$\mathrm{NiSO}_4\cdot 6\mathrm{H}_2\mathrm{O}$

14.

NaCl

15.

K⁺

16.

CH₄

17.

HNO₃

Balancing chemical reactions

1.

$2\mathrm{NaCl} + \mathrm{SO_2} + \mathrm{H_2O} + \frac{1}{2}\mathrm{O_2} \rightarrow \mathrm{Na_2SO_4} + 2\mathrm{HCl}$

2.

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + 3\mathrm{O_2} \rightarrow 2\mathrm{CO_2} + 3\mathrm{H_2O}$

Basic physical chemistry

1.

In SI units, the data are as follows: $V=75\,\mathrm{m^3}$, $P = 90\,000\,\mathrm{Pa}$ and $T = 298\,\mathrm{K}$. The number of moles is therefore

$$n = \frac{PV}{RT} = \frac{(90\,000\,\mathrm{Pa})(75\,\mathrm{... ...m{J\,K^{-1}mol^{-1}})(293\,\mathrm{K})} = 2.77\,\mathrm{kmol}.$$

2.

$q = m\tilde{C}_P\Delta T = (200\,\mathrm{g})(4.184\,\mathrm{J\,K^{-1}g^{-1}})(16\,\mathrm{K}) = 13\,\mathrm{kJ}$

3.

$q = m\Delta\tilde{H}^\circ_{(\mathrm{condense},100^\circ\mathrm{C})} = (400\,\mathrm{g})(-2257\,\mathrm{J/g}) = -903\,\mathrm{kJ}$

4.

$$\begin{eqnarray*}\Delta\bar{H}^\circ & = & \Delta\bar{H}^\circ_{f(\mathrm{BF_3})... ...}(0)\right)\,\mathrm{kJ/mol}\\ & = & -732.2\,\mathrm{kJ/mol} \end{eqnarray*}$$

5.

$$K = \frac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]}$$

6.

You first have to look up the acid dissociation constant of acetic acid. The values vary somewhat from table to table. I got mine from Kotz and Treichel, Chemistry & Chemical Reactivity, 3rd ed., p 808. According to this source, $K_a = 1.8\times 10^{-5}$. This is the equilibrium constant for the reaction

$$\mathrm{CH_3COOH_{(aq)}} \rightarrow \mathrm{CH_3COO^-_{(aq)}} + \mathrm{H^+_{(aq)}}.$$

Thus

$$K_a = \frac{[\mathrm{CH_3COO^-_{(aq)}}][\mathrm{H^+_{(aq)}}]}{[\mathrm{CH_3COOH_{(aq)}}]}.$$

There are two ways to proceed:

(a)

We have a total of $0.01\,\mathrm{mol/L}$ of acetic acid. After dissociation, this means that

$$[\mathrm{CH_3COO^-_{(aq)}}]+ [\mathrm{CH_3COOH_{(aq)}}] = 0.01\,\mathrm{mol/L}.$$

Moreover, one proton is produced for each acetate ion. We can neglect the water equilibrium since the acid concentration is significant and the K_a relatively large. Accordingly, $[\mathrm{CH_3COO^-_{(aq)}}] \approx [\mathrm{H^+_{(aq)}}]$. Combining these equations, we obtain a quadratic equation. Working through this procedure, I find $[\mathrm{H^+_{(aq)}}] = 4.15\times 10^{-4}$. I'm not showing the detailed work because there is a simpler way which you should use whenever possible.

(b)

The K_a is not particularly large. It follows that a relatively small amount of acid will dissociate so that $[\mathrm{CH_3COOH_{(aq)}}] \approx 0.01\,\mathrm{mol/L}$. The validity of this approximation can be verified at the end of the calculation. It is still true that $[\mathrm{CH_3COO^-_{(aq)}}] \approx [\mathrm{H^+_{(aq)}}]$. The K_a equation becomes

$$\begin{eqnarray*}% latex2html id marker 122 1.8\times 10^{-5} & \approx & \frac... ...therefore [\mathrm{H^+_{(aq)}}] & \approx & 4.24\times 10^{-4}. \end{eqnarray*}$$

Note that the answer justifies our approximations. It is quite a bit smaller than the total amount of acetic acid and quite a bit larger than the proton concentration which might be obtained due to the dissociation of water. The difference between the answers obtained by the two methods is negligible on the logarithmic pH scale.

The pH is therefore 3.4.

Note that I didn't carry around units in these calculations. There is a reason for this which we will study later in this course.

7.

(a)

no

(b)

yes

(c)

no

(d)

yes

(e)

yes

8.

(a)

no

(b)

yes

(c)

no

(d)

yes

(e)

no

Mathematics

1.

$$\frac{df}{dx} = 15x^4 + 2$$

2.

(a)

$$\int f(x)\,dx = \frac{3}{6}x^6 + \frac{2}{2}x^2 + x = \frac{1}{2}x^6 + x^2 + x$$

(b)

$$\int_{x_i}^{x_f}\frac{dx}{x} = \left.\ln x\right\vert _{x_i}^{x_f} = \ln x_f - \ln x_i = \ln\left(\frac{x_f}{x_i}\right)$$

(c)

$$\begin{eqnarray*}\displaystyle\int_1^2\frac{3x^2+8x+1}{2x}\,dx & = & \int_1^2 d... ...\ & = & \left(\frac{25}{4} + \frac{1}{2}\ln 2\right) = 6.5966 \end{eqnarray*}$$