Calculate the concentration of the constituentions in solutions of the following compounds in the following concentrations: (a) 0.45 M HBr, (b) 0.045 M KOH, (c) 0.0112 M CaCl2.
(a) 0.45 M HBr:
• Hydrogen ion (H+) concentration: 0.45 M
• Bromide ion (Br-) concentration: 0.45 M
(b) 0.045 M KOH:
• Hydroxide ion (OH-) concentration: 0.045 M
• Potassium ion (K+) concentration: 0.045 M
(c) 0.0112 M CaCl2:
• Calcium ion (Ca2+) concentration: 0.0112 M
• Chloride ion (Cl-) concentration: 0.0224 M
To calculate the concentration of the constituents in the given solutions, we need to know the dissociation behavior of each compound.
(a) 0.45 M HBr:
HBr is a strong acid and dissociates completely in water. Therefore, it will produce one hydrogen ion (H+) and one bromide ion (Br-).
The concentration of H+ ions is the same as the initial concentration of HBr, which is 0.45 M.
The concentration of Br- ions is also 0.45 M.
(b) 0.045 M KOH:
KOH is a strong base and dissociates completely in water. This results in one potassium ion (K+) and one hydroxide ion (OH-).
The concentration of K+ ions is the same as the initial concentration of KOH, which is 0.045 M.
The concentration of OH- ions is also 0.045 M.
(c) 0.0112 M CaCl2:
CaCl2 dissociates into one calcium ion (Ca2+) and two chloride ions (Cl-).
The concentration of Ca2+ ions is half of the initial concentration of CaCl2, which is 0.0112 M / 2 = 0.0056 M.
The concentration of Cl- ions is twice the initial concentration of CaCl2, which is 0.0112 M * 2 = 0.0224 M.
So, the concentration of constituent ions in the given solutions is as follows:
(a) 0.45 M H+ and 0.45 M Br-
(b) 0.045 M K+ and 0.045 M OH-
(c) 0.0056 M Ca2+ and 0.0224 M Cl-
To calculate the concentration of the constituents in a given compound, you need to understand the chemical formula of the compound and the stoichiometry of the reaction between the compound and water.
(a) 0.45 M HBr:
HBr is a binary acid, which means it ionizes in water to produce hydrogen ions (H+) and bromide ions (Br-). HBr is a strong acid, so it fully ionizes in water.
The concentration of hydrogen ions (H+) in the solution is equal to the concentration of the acid itself. Therefore, the concentration of H+ in 0.45 M HBr is 0.45 M.
The concentration of bromide ions (Br-) is also 0.45 M since one mole of HBr forms one mole of Br- ions upon ionization.
(b) 0.045 M KOH:
KOH is a strong base that fully dissociates in water to produce potassium ions (K+) and hydroxide ions (OH-).
The concentration of hydroxide ions (OH-) in the solution is equal to the concentration of the base itself. Therefore, the concentration of OH- in 0.045 M KOH is 0.045 M.
The concentration of potassium ions (K+) is also 0.045 M since one mole of KOH forms one mole of K+ ions upon dissociation.
(c) 0.0112 M CaCl2:
CaCl2 is an ionic compound. When it dissolves in water, it dissociates into one calcium ion (Ca2+) and two chloride ions (Cl-).
The concentration of calcium ions (Ca2+) in the solution is half the concentration of the compound itself since one mole of CaCl2 forms one mole of Ca2+ ions upon dissociation. Therefore, the concentration of Ca2+ in 0.0112 M CaCl2 is 0.0056 M.
The concentration of chloride ions (Cl-) is twice the concentration of the compound itself since one mole of CaCl2 forms two moles of Cl- ions upon dissociation. Therefore, the concentration of Cl- in 0.0112 M CaCl2 is 0.0224 M.
Remember that these calculations assume complete dissociation of the compounds in water.