Expressing
concentration
concentration
If 0.750g of methanol is dissolved in 12.0mL of the solution, What is the %w/v?
(0.750g/ 12.0 mL )X 100%= 6.25% w/v
Sucrose is added to water in a solution. C12H22O11(s) What happens to Sucrose when added into water and what is the state?
potassium hydroxide is added into a solution, what happens to the potassium hydroxide and what are the states?
C12H22O11(s)--> C12H22O11(aq) ( molecular) highly soluble, undergoes full dispersion.
What happens to the solubility of solids AND solubility gases as temperature increases?
As temperature increases the solubility of solids increases. As temperature increases, the solubility of gases decreases.
The pH of human blood is measured to be 7.40. What is the concentration of the hydronium ion in the blood?
[H3O+(aq)] = 10-pH = 10-7.40 = 3.9 x 10-8 mol/L
Which of the acids listed below are polyprotic (meaning they will undergo more than one reaction when placed in water)?
H2SO4(aq), HCl(aq), CH3COOH (aq) , H3PO4(aq)
Answer: H2SO4(aq), H3PO4(aq)
A sample of water is analyzed in a lab and it is found that the sample contains 50.0 ppm of pesticides. The mass of the pesticides dissolved in 750 mL of solution is ____?
50.00 ppm= 50.0mg/ 1L
m=cV= 50.0 mg/L X 0.750 L
Answer = 37.5 mg
Non-polar liquids are _________ in water and have increased solubility in water at a __________ temperature. Which of the following four rows completes the statement above?
A.) immiscible lower
B.) miscible higher
C.) miscible lower
D.) immiscible higher
D.) immiscible higher
Calculate the mass of silver nitrate needed to prepare 2.00 litres of 0.325 mol/L silver nitrate solution.
m=? M = 169.88 g/mol V = 2.00 L c = 0.325 mol/L
n=cV = 0.325 mol/L x 2.00 L = 0.650 mol
m = nM = 0.650 mol x 169.88 g/mol
m = 110 g
The pH value of an aqueous solution having a hydronium ion concentration equal to 3.0 x 10-12 mol/L is _________ and the solution would be classified as a _________
pH = 11.52 and the solution is a base
Samples of NH3(aq), HBr(aq), H2SO3(aq), and KOH(aq) were accidentally mixed up in the lab. Diagnostic tests were done so that the samples could be properly labelled. Use the lab results to determine the correct identity of each sample. ( Slide 26)
Answer: 1 = HBr, 2 = KOH, 3 = H2SO3, 4 = NH3
Calculate the concentration as a % W/V if 4.5 g of calcium chloride is dissolved in 50.0 mL of solution.
c=4.5g/50.0 mL X 100= 9.0% w/V
Draw a beaker with particles representing the entities that would be found if a weak base such as sodium hydrogen sulphite was dissolved in water. Begin by writing an ionization equation.
HSO3- (aq) + H2O(l) ↔ OH- (aq) + H2SO3(aq)
slide 10.
C1= 8.00 mol/L V1=?
c2= 0.500 mol/L V2= 0.200 L
c1v1=c2v2
v1= (c2v2)/ C1
(0.500 mol/L X 0.200 L) / ( 8.00 mol/L) = 0.0125 L or 12.5 mL
Calculate the [H3O+(aq)] of a sample, if it turns orange when thymol blue indicator is added.
Indicators can be used to estimate the pH of a sample.
Thymol blue will be orange if the pH falls between 1.2 and 2.8. The average of these two numbers (the approximate pH) is 2.0. Use this average to calculate the [H3O+(aq)]
[H3O+(aq)] = 10 –pH = 10 -2.0
[H3O+(aq)] = 1 x 10-2 mol/L
Write a chemical reaction equation that explains the acidic properties of formic acid (HCOOH(aq))
HCOOH (aq) + H2O (l) <--> H3O+(aq) + HCOO- (aq)
A farmer dissolves 50 g of urea, (NH2CONH2), in 5.0 L of water to spray her fields. What is the amount concentration (mol/L) of the solution?
m= 50 g M= 60.07 g/mol V=5.0 L
n=m/M = 50g/ (60.07g/mol)= 0.83236 ...mols
c=n/v = 0.83236 ...mols/5.0L =0.17mol/L
Write the dissociation equation and calculate the concentration of the ions in a 3.00 mol/L cobalt(III) acetate solution. Remember to use the mole ratio.
Co(CH3COO)3(aq) --> Co3+(aq)+ 3 CH3COO-(aq)
C= 3.00 mol/L C= 3.00 mol/L C= 9.00 mol/L
A 100 mL solution is carefully prepared from a known mass of solid solute. Place in order, the following four steps taken from the lab procedure.
1.) Transfer to a 100 mL volumetric flask using a funnel
2.) Add solvent to the calibration line on the volumetric flask using an eyedropper for the last few mLs.
3.) Dissolve the solid in approximately 40-50 mL of water in a 100 mL beaker
4.) Rinse glassware used for dissolving three times and pour the rinse water into the volumetric flask
ANSWER: 3 1 4 2
A technician tested three samples of the same solution with three different indicators and the following results were obtained.
Methyl orange = orange
Orange IV = yellow
Bromocresol green = yellow
Estimate the approximate pH of the solution.
Answer: pH is between 3.2 – 3.8
Identify the liquids from the following list that would be miscible (would dissolve in all proportions) in water.
benzene(C6H6(l)), methanol (CH3OH(l)), methanoic acid (HCOOH(l)), octane (C8H18(l))
ANSWER: methanol and methanoic acid will be miscible
What is the % W/V of a 2.50 mol/L hydrogen peroxide, (H2O2(aq)) solution?
Note: You need to convert from mol/L to g/mL then multiply by 100 to get a % for 100 mL.
M of H2O2= 34.02 g/mol
2.50 mol/L X 1 L/ 1000 mL X 34.02 g/ 1mol X 100
Answer is 8.51% w/v
The pH of a solution is raised from 7-10. What happens to the [H3O+(aq)]? How many steps and factors did it change by?
The pH is increased by 3 steps, 7-->8 8-->9 9-->10.
[H3O+(aq)] is decreased by 3 x “factors of 10” (10 x 10 x 10). In other words, the [H3O+(aq)] is 1000 times smaller. Or in other words 10^3= 1000 times less or decreased.
A saturated salt solution is prepared for a Chem 20 lab. Although the solution is shaken before it is used in the lab, solid salt crystals remain at the bottom of the volumetric flask. Explain the theory of dynamic equilibrium as it relates to the presence of solid crystals at the bottom of the saturated solution.
It is saturated when no more solute will dissolve.
(at that particular temperature)
A saturated solution is a dynamic equilibrium because dissolving and crystallizing occur at the same rate.
The pOH of a solution obtained by dissolving 7.00 g of NH4OH(s) in 400 mL of water is ____________.
NH4OH --> NH4+(aq) + OH-(aq)
= Complete dissociation in a 1:1 mol ratio So [NH4OH(aq)] = [OH-(aq)]
n=m/M = 7.00 g / 35.06 g/mol =0.19965777.... mol
c=n/V =0.19965777.... mol/ 0.400 L =
Answer is 0.499 mol/L
Solutions of nitric acid and oxalic acid both have a concentration of 0.25 mol/L. Explain why the pH of the nitric acid is lower than the pH of the oxalic acid at the same temperature.
Nitric acid is a strong acid that fully ionizes. Each molecule of acid reacts with water to form H3O+ (aq). Increased hydronium ions means a lower pH.
Oxalic acid is a weak acid. Only some of the acid molecules react with water and therefore there are fewer hydronium ions. When hydronium ion concentration is reduced, pH is increased.