Extra Credit 3

Last updated
Save as PDF

Page ID: 96914

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

Q5.3.5

How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat.

Solution:

The heat transferred by a substance is determined as the equation:

$$q = Cm \Delta T $$

Assuming there is no heat loss in the process of heat transfer, the heat absorbed by water at 25 °C should equal the heat released by water at 95 °C, thus:

$$q_1 = C m_1 \Delta T_1= C m_2 \Delta T_2 = q_2 $$

Where m₁ is the mass of water added, $\Delta T_1$ is the temperature change of water added, m₂ is the mass of the coffee and $\Delta T_2$ is the temperature change of coffee. Here, we also assume the specific heat capacity of coffee is equal to water. The specific heat of water here is 4.18 J/g·°C. There is only one unknown, which is m₁, and we can substitute other values in to this equation and get the value we need.

Answer

$\ce{(1.{7})*10^2}mL{H_2O}$

*decimal 1.7

Q14.3.58

Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 M in NH₃ and 0.100 M in C₆H₅NH₂.

Solution:

First we can write out the two chemical equations: $$NH_3(aq) +H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$$

$$ C_6H_5NH_2(aq) + H_2O(l) \rightleftharpoons C_6H_5NH_3^+(aq) + OH^-(aq)$$

And we can combine these two equations:

$$ NH_3(aq) + C_6H_5NH_2(aq) + 2H_2O(l) \rightleftharpoons NH_4^+(aq) +C_6H_5NH_3^+(aq)+2OH^-(aq)$$

We know that:

$$K_{b(NH_3)} = \frac {[NH_4^+][OH^-]}{[NH_3]} $$

$$K_{b(C_6H_5NH_2)} = \frac {[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]} $$

The equilibrium constant $K_{rxn}$ of the combined reaction is $$K_{rxn} = \frac {[C_6H_5NH_3^+][NH_4^+][OH^-]^2}{[C_6H_5NH_2][NH_3]} = K_{b(NH_3)}·K_{b(C_6H_5NH_2)}$$

Based on the equilibrium constant above and reaction equation above, we can set up an ICE table for this reaction, then we can use the ice table to get the equilibrium concentration of unionized base for both species.

Answer

$C_{10}H_{14}N_2$= 0.049 M

$C_{10}H_{14}N_2H^+$ = 1.9 × 10−4M {

$C_{10}H_{14}N_2H_{22}^+$ = 1.4 × 10−11M

$OH^-$= 1.9 × 10−4M
$H_3O^+$= 5.3 × 10−11M

Q14.1.6

Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:

HS⁻
$\ce{PO4^3-}$
$\ce{NH2-}$
C₂H₅OH
O²⁻
$\ce{H2PO4-}$

Solution:

Based on the definition of Brønsted-Lowry base, we have to write an equation for these substances above to accept protons. For example, for the first species, it can accept protons as the equation below:

$$H^+(aq) + HS^- (aq)\rightleftharpoons H_2S(aq)$$

The basic concept here is how to express these species in equation to show how they accept protons.

a. $\ce{HS^-}$+$\ce{H_2O}$$\rightleftharpoons\ce{H_2S}$+$\ce{OH-}$

b. $\ce{PO_4^3-}$+$\ce{H_2O}$$\rightleftharpoons\ce{HPO_4^{2-}}$+$\ce{OH-}$

c. $\ce{NH_2^-}$+$\ce{H_2O}$$\rightleftharpoons\ce{NH_3}$+$\ce{OH-}$

d. $\ce{C_2H_5OH}$+$\ce{H_2O}$$\rightleftharpoons\ce{C_2H_5OH_2^+}$+$\ce{OH-}$

e. $\ce{O^{2-}}$+$\ce{H_2O}$$\rightleftharpoons\ce{OH-}$+$\ce{OH-}$

f. $\ce{HPO_4^{2-}}$+$\ce{H_2O}$$\rightleftharpoons\ce{H_2PO_4^-}$+$\ce{OH-}$

Q15.1.16

Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:

Ag₂SO₄
PbBr₂
AgI
CaC₂O₄•H₂O

Solution:

In these questions, the major problem is how to use $K_{sp}$ to deal with the solubility. To solve this, we have to write out the composition of $K_{sp}$ to find which spices can be used to determine the solubility of the substance. For example, for AgI, it has the expression of $K_{sp}$:

$$ K_{sp} = [Ag^+][I^-] $$

In this case, either $\ce {[Ag^+]}$ or $\ce {[I^-]}$ can represent the solubility of AgI because there are 1 mol of Ag⁺ and I^- in 1 mol of AgI. Because there is no other ion species affecting $K_{sp}$, the value of $\ce {[Ag^+]}$ or $\ce {[I^-]}$ can be calculated by:

$$ [Ag^+] = [I^-] = \sqrt {K_{sp}} $$

The result is the solubility of $\ce {AgI}$.

Differently, for $\ce {PbBr_2}$,

$$ K_{sp} = [Pb^{2+}][Br^-]^2 $$

In this case, we can use $\ce {[Pb^+]}$ to determine the concentration of $\ce {PbBr_2}$, which can be calculated by:

$$ [Pb^{2+}] = \sqrt {\left(\frac{K_{sp}}{[Br^-]^2} \right)}$$

Where

$$ 2[Br^-] = [Pb^{2+}] $$

Answer

a. $\ce {4.49}(g/L)$

b. $\ce {4.26}(g/L)$

c. $\ce {(1.2)*10^{5-}}(mol/L)$

d. $\ce {(4.8)*10^{5-}}(mol/L)$

dot is a decimal not *

Q16.2.7

Arrange the following sets of systems in order of increasing entropy. Assume one mole of each substance and the same temperature for each member of a set.

H₂(g), HBrO₄(g), HBr(g)
H₂O(l), H₂O(g), H₂O(s)
He(g), Cl₂(g), P₄(g)

Solution:

There are three principles to compare the entropies of different systems:

1. For substances that have the same composition but different states: $$S_{gas} > S_{liquid} > S_{solid}$$

2. For molecules that in the same state, the one has more atoms has greater entropy.

3. For molecules that have the same composition in the same state, the softer one has greater entropy.

Answer

a. $\ce{H_2}(g)$, $\ce{HBr}(g)$, $\ce{HBrO_4}(g)$

b. $\ce{H_2O}(s)$, $\ce{H_2O}(l)$, $\ce{H_2O}(g)$

c. $\ce{He}(g)$, $\ce{Cl_2}(g)$, $\ce{P_4}(g)$