Skip to main content
Chemistry LibreTexts

Homework Solutions #3

  • Page ID
    36923
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    3.1

    Write the products of these acid-base reactions. Balance the complete reaction.

    1. \(HF_{(aq)} + KOH_{(aq)} \rightarrow\)
    2. \(H_2SO_{4\;(aq)} + Ca(OH)_{2\;(aq)} \rightarrow\)

    A4.132Z

    1. \(HF_{(aq)} + KOH_{(aq)} \rightarrow KF_{(aq)} + H_2O_{(l)}\)
    2. \(H_2SO_{4\;(aq)} + Ca(OH)_{2\;(aq)} \rightarrow CaSO_{4\;(s)} + 2H_2O_{(l)}\)

    H4.132Z

    S4.132Z

    Acid + Base -> Salt + Water

    1. \(K^+\) is soluble with \(F^-\). Equation is already balanced.
    2. \(Ca^{2+}\) is insoluble with \(SO_4^{2−}\). Balance the equation.

    3.2

    Balance the following redox reaction in basic conditions with state symbols.

    \[H_2O_{(l)} + Na_{(s)} \rightarrow Na^+_{(aq)} + H_{2\;(g)}\]

    A4.88V

    \[2H_2O_{(l)} + 2Na_{(s)} \rightarrow 2Na^+_{(aq)} + 2OH^-_{(aq)} + H_{2\;(g)}\]

    H4.88V

    S4.88V

    • \(H\)is reduced from +1 to 0;
    • \(Na\) is oxidized from 0 to +1.

    The H in \(NaOH\) and the \(H\) in \(H_2O\) are both +1. The \(H_2\) is the reduced product and \(NaOH\) is oxidized product. Since the change is both 1, the number of H in \(H_2\) should be the same as the number of Na in \(NaOH\). There are 2 H in \(H_2\), there should be 2 \(NaOH\).

    +1 0 +1 0

    H2O(l) + Na(s) → NaOH(aq) + H2(g)

    3.3

    n a lab experiment, a 20.00 ml \(HNO_2\) solution was titrated with \(0.1234\; M\) sodium hydroxide solution, and the equivalence point was reached when 43.21 ml of this basic solution was added. What is the original molarity of the \(HNO_2\) solution?

    H4.74O

    A4.74O

    • 0.2667M \(HNO_2\) solution.

    S4.74O

    The net ionic equation for the reaction is

    \[HNO_{2\; (aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)} + NO^-_{2\; (aq)}\]

    Be careful here. Weak acid and should not be separated when writing in the net ionic equation, so neither of its conjugate base on both sides should be canceled out. The exact amount of hydroxides requires to balance protons and form water is 1:1, and this particular point is called equivalence point. The amount of mole= molarity x volume (L).

    The amount of hydroxide:

    \[0.1234\;M \times 0.04321\;L = 0.005332\; mol \;OH^-\]

    Same amount of \(H^+\) was titrated in 20.00 ml, so we can calculate the molarity:

    \[\dfrac{0.005332 \; mol\; H^+}{0.02000 \; L} = 0.2667\; M\]

    3.4

    What is the required mL of 1.25 M \(HBr\) necessary to completely neutralize a 175 mL sample of 2.38 M \(Mg(OH)_2\)?

    A4.117B:

    • 666 mL

    H4.117B:

    S4.117B:

    Multiply the Molarity of \(Mg(OH)_2\) by its volume, then multiply it again by two to account for the two moles of \(OH^-\) present per mole of \(Mg(OH)_2\). An equal amount of acid neutralizes this, so an equal amount of moles of \(HBr\) will be needed. Divide this by the molarity to find the volume necessary.

    3.5

    A rubber duck is initially filled with 100 mL air at \(2 ^oC\) in the bathroom and then set in the sun to be heated. The final temperature increases to \(10 ^oC\) at constant pressure.

    What is the final condition of the rubber duck? Does it shrink or expand? What is its final volume?

    A5.44P

    The rubber duck expands. \(V_f=103\;mL\)

    H5.44P

    S5.44P

    Step 1: Write down your given information:

    • Pressure is constant: \(P_i=P_f\)
    • Volume changes: \(V_i=100\;mL\) and \(V_g = unknown\)
    • number of moles is constant: \(n_i = n_f\)
    • Temperature changes: \(T_i=2^oC\) and \(T_f=10^oC\)
    • \(R=0.0820574 L \times atm \times mol^{-1}K^{-1}\)

    Step 2: Convert as necessary:

    • Temperature:

    \[T_i=2^oC+ 273\; K= 275\; K\]

    \[T_f=10^oC+273\;K=283\; K\]

    Step 3: Plug in the variables into the appropriate equation:

    \[V=\dfrac{nRT}{p}\]

    \[\dfrac{V}{T}=\dfrac{nR}{P}\]

    \[\dfrac{100\;mL}{275\;K}=\dfrac{nR}{P}\]

    \[V_f=\dfrac{nRT_f}{P_f}\]

    \[V_f=\dfrac{T_f\;nR}{P_f}\]

    \[V_f=\dfrac{283\;K \times 100\; mL}{275\;K}\]

    \[\therefore \; V_f=103\;mL\]

    The duck expands, but barely.

    3.6

    A commercial balloon is filled with 3000 L helium gas at \(30^oC\). If the pressure is 1 atm, what is the mass of helium gas? Why does the balloon float?

    A5.51P

    • 484 g

    H3.51P

    S5.51P

    Step 1:Write down your given information:

    • \(P=1\;atm\)
    • \(V=3000 \;L\)
    • \(n =?\)
    • \(T=30 ^oC\)
    • \(R=0.0820574 \;L \times atm \times mol^{-1}K^{-1}\)

    Step 2: Convert as necessary:

    • Temperature:

    \[30^oC+273\;K=303\;K\]

    • Pressure: No need since R is given in terms of \(atm\)

    Step 3: Plug in the variables into the appropriate equation:

    \[V=\dfrac{nRT}{p}\]

    \[n=\dfrac{Vp}{RT}\]

    \[n=\dfrac{3000\;L \times 1\; atm}{0.0820574 \;L \times atm \times mol^{-1}K^{-1} \times 303\;K}\]

    \[n=121\; moles\]

    Now convert to mass via molecular weight of \(He\) (taken from periodic table)

    \[mass=n \times \text{molar mass}= 121\; moles \times 4.002\; g/mol= 484\; g\]

    While the balloon has a mass and hence attracted to the earth via gravity, Archimedes' principle indicates that an upward buoyant force is equal to the weight of the fluid that the body displaces, which is air (\(N_2\)). Because \(N_2\) is heavier than \(He\), the ballow floats.

    3.7

    Uranium hexafluoride (\(UF_6\)) is a compound used in the uranium enrichment process that produces fuel for nuclear reactors and nuclear weapons. It forms solid grey crystals at standard temperature and pressure and boils at 56°C.

    Figure: Uranium hexafluoride crystals sealed in an ampoule

    The density of uranium hexafluoride is 12.6 g/L at 745 torr. What is the temperature in °C under these condition?

    A.5.79W:

    • 60.°C

    H5.79W:

    S.5.79W:

    We can derive a new formula about density from \(PV=nRT\):

    Solve the equation for temperature

    \[T= \dfrac{PV}{nR} \tag{1}\]

    The number of moles is the place to start to find the mass. The number of moles is the mass (\(m\)) of the gas divided by its molecular mass (\(M\)):

    \[n=\dfrac{m}{M} \tag{2}\]

    when substituted into equation 1, gives

    \[T=\dfrac{PV}{nR}= \dfrac{PVM}{mR} \tag{3}\]

    Density is defined as mass/Volume

    \[\rho= \dfrac{m}{V} \tag{4a}\]

    we rearrange equation 4a for \(V\)

    \[V= \dfrac{m}{\rho} \tag{4b}\]

    and substitute equation 4b into equation 3

    \[T= \dfrac{P \dfrac{m}{\rho}M}{mR} \tag{5}\]

    After simplify the equation, we have

    \[T= \dfrac{M\;P}{\rho R}\]

    \[T= \dfrac{M\;P}{\rho R} = \dfrac{(352.05\; g/mol)(745\; torr)(1 \; atm/760\; torr)}{(12.6\; g/L)/( 0.08206 \; L \; atm/K mol)}=333\; K\]

    \[T= 333 \;K -273\; K=60 ^oC\]

    3.8

    A constant-volume container has 2.00 moles gas sample at 25 °C and the gas sample exerts a pressure of 450 torr. If 2.00 moles more gas is added in the container and the temperature is increased to 50 °C, what is the pressure in atm?

    A5.55W:

    • 1.28 atm

    H5.55W: Find out the given value \(n_1\) and \(n_2\). \(T_1\) and \(T_2\), Be careful of the data for \(n_2\) and the unit of temperature. We use Kelvin here for the temperature, so we need to convert the Celsius to Kelvin with T=T’+273K. Use \(PV=nRT\) to write two equations and then calculate the pressure.

    S5.55W:

    Ggiven

    • \(n_1=2.00\;moles\)
    • \(T_1= 25+273\; K=298\;K\)
    • \(P_1= 450\;torr=0.59\;atm\)
    • \(n_2 = 4.00\; moles\)
    • \(T_2= 50+273=323\;K\)

    Want to find \(P_2\).

    According to PV=nRT, P1V1=n1RT1 P2V2=n2RT2

    Since the volume is constant,

    \[\dfrac{P_1}{P_2}= \dfrac{n_1T_1}{n_2T_2}\]

    \[P_2= \dfrac{P_1n_2T_2}{n_1T_1}= \dfrac{(0.59)(4)(323)}{(2 \times 298}= 1.28\; atm\]

    We can see that although the \(n_2\) and the \(T_2\) in Celsius are twice as \(n_1\) and \(T_1\) in Celsius, we cannot say that \(P_2=2P_1\), which will give us a wrong answer.


    Homework Solutions #3 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?