Conceptual Questions
#1. Explain the concept of "Like Dissolves Like." What kind of substances are dissolvable in water according to this principle?
#2. What happens if you continue to add solute into a solution that is at its solubility limit?
#3. Define acids and bases according to both the Arrhenius and Bronsted-Lowry definitions.
#4. What is the significance of the pH scale being a logarithmic scale?
#5. What is the difference between a strong acid and a weak acid? A strong base and a weak base?
#6. What are the products of acid-base reactions?
#7. Describe what an oxidation-reduction reaction is. How do you determine the oxidizing and reducing agents in an equation?
Practice Questions
#1. Calculate the molarity of a solution with `2.7 x 10^(24)` atoms `Li` in 2L of solution.
#2. Calculate the molarity of all ions for 150g of `CaCl_2` in 3L of solution.
#3. Calculate the pH of a solution of 0.1M `H_2SO_4`.
#4. You find a beaker filled with 200mL of an unknown base. It takes 50mL of 6M HCl to neutralize the base. What is the concentration of the base?
#5. Write out the complete and net ionic equations of the reaction:
`Mg(NO_3)_(2 (aq)) + Na_2CO_(3 (aq)) rArr MgCO_(3 (s)) + 2NaNO_(3 (aq))`
#6. Determine the oxidation states of all atoms in the molecule `K_2C_2O_4`.
#7. Balance the following redox reaction:
`Au^(3+) + I^- rArr Au + I_2`
Answers
Conceptual Questions
#1. The concept describes how polar substances are likely to dissolve in polar solutions whereas non-polar substances are likely to dissolve in non-polar solutions. Water will dissolve most polar substances because water is a polar solution
#2. The solubility limit is defined as the limit to how much solute can be dissolved in a particular solvent. Once this limit is reached, no more of the solute will dissolve. Instead, the solute will just remain as a solid dispersed throughout the solution.
#3. According to the Arrhenius definition, acids release `H^+` in water while bases release `OH^-`. According to the Bronsted-Lowry definition, acids are proton donors while bases are proton acceptors.
#4. Since pH is a logarithmic scale, each whole number increase in pH corresponds to a 10x increase in concentration of `H^+` ions. A solution that is pH 2 is 10x more acidic than a solution that's pH 3.
#5. Strong acids and bases dissociate fully to their ions whereas weak acids and bases will only dissociate partially.
#6. An acid-base reaction will usually result in a salt and water.
#7. An oxidation-reduction reaction is a reaction that involves a transfer of electrons. You can determine the oxidizing and reducing agents in an equation by looking at what's being oxidized and what's being reduced.
Practice Questions
#1. First, convert to moles:
`(2.7 x 10^24 "atoms")((1 mol")/(6.022 x 10^23 "atoms"))= 4.5 "moles Li"`
Now we can convert to molarity by using `M="moles"/L`:
`M= (4.5 "moles Li")/(2L)= 2.25M`
#2. Notice that the substance is an ionic solid. This means that it's going to dissociate, with different molarities for the different ions. First, write out the dissociation:
`CaCl_2 rArr Ca^(2+) + 2Cl^-`
Now we can calculate the molarity of each individual ion. Convert 150g `CaCl_2` to moles `CaCl_2`:
`(150 g CaCl_2)((1" mol")/(111 g CaCl_2))= 1.4" mol" CaCl_2`
Since the `Ca^(2+)` ion is in a 1:1 ratio with `CaCl_2`, the moles of `Ca^(2+)` will be the same as the moles `CaCl_2`. On the other hand, there will be twice as many moles of `Cl^-`.
`M_(Ca^(2+))=(1.4" mol" Ca^(2+))/(3 L)=0.47M Ca^(2+)`
`M_(Cl^-)=(2.8" mol" Cl^-)/(3L)=0.83M Cl^-`
Notice that the `M_(Cl^-)` is exactly twice as much as the `M_(Ca^(2+))`
#3. pH= -log[`H^+`]. We just need the [`H^+`]. Seeing as `H_2SO_4` is a strong acid, we can assume that all of it will dissociate. Therefore, [`H_2SO_4`]=[`H^+`]=0.1M. The pH is -log[0.1]= 1. That is a pretty acidic solution.
#4. Use `M_1V_1=M_2V_2`
`(M_1V_1)/V_2=M_2`
`((6M H^+)(0.05 L))/(0.2 L)=1.5M "base"`
#5. Complete ionic equation (Remember that `NO_3^-` and `CO_3^(2-)` are polyatomic ions):
`Mg_((aq))^(2+) +2NO_(3 (aq))^- 2Na_((aq))^+ +CO_(3 (aq))^(2-) rArr MgCO_(3 (s)) +2Na_((aq))^+ + 2NO_(3 (aq))^-`
Net ionic equation:
`Mg_((aq))^(2+) + CO_(3 (aq))^(2-) rArr MgCO_(3 (s))`
#6. `Q_K=+1, Q_C=+3, Q_O=-2`
#7. Write out the half reactions and balance them:
`Au^(3+) + 3e^- rArr Au`
`I^- rArr I_2 + e^-`
`Au^(3+) + 3e^- rArr Au`
`6I^- rArr 3I_2 + 3e^-`
Now combine the half-reactions into one equation:
`Au^(3+) + 6I^- rArr Au + 3I_2`