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How to Find Molar Concentration of Ions
The solution is a homogenous mixture of solute in solvent. The solute is that component which is found in small amount compare to other component that is solvent.
Solvent is found in high concentration in a solution. The nature of a solution depends upon the solute and solvent.
It also depends upon the concentration of solution and chemical nature of solute and solvent. The concentration of a solution represents the amount of solute present in a certain amount of solvent.
There are several ways to represent the concentration of a solution such as molarity,normality,ppm,formality,percentage composition etc.
Let’s discuss one of the ways to express the concentrations of solutions quantitatively which is known as molarity of the solution.
It is one of the most widely used quantitative relations to express the concentration of solution.
It can be defined as the number of moles of solute present in one liter of solution. The mathematical relation of molarity with the number of moles of solute can be written as given below;
Molarity (M) = Number of moles of solute (n) / Volume of solution in liters (V)
Hence, one molar solution will contain 1.00 mol of solute in one liter of the solution.
Let’s take an example of molarity. Find the molarity of a solution which is made by dissolving 23.4 g of sodium sulphate (Na_{2}SO_{4}) to form 125 mL of solution.
The molar mass of sodium sulphate is 142 gm; hence we can calculate the moles of sodium sulphate would be 23.4 gm / 142 gm/mol = 0.165 mol of Na_{2}SO_{4}. The volume of solution is 125 ml or 0.125 L; hence the molarity of the solution would be;
Molarity = 0.175 mol / 0.125 = 1.32 M
We can calculate the concentration of free ions found in the solution of ionic compounds.
The use of mole ratio is the best way to calculate the concentration of ions in the solution.
We know that anions are negatively charged ions while cations are positively charged ions
The dissociation equation for a certain compound can be written as given below;
xA > yB + zC
Here A stands for the substance is being dissolved before while B and C are ionic forms in the equation. x,y, and z represents the coefficients of ions involve in the reaction.
To calculate the concentration of ion with the help of mole ratio; let’s start with same reaction; the concentration of solute is [A], concentration of 1st ion is [B] or y/x [A].
The concentration of 2nd ion is [C] or z/x [A]. For example; calculate the ion concentration of 0.050 mol/L solution of KCl.
KCl (s) > K+ (aq) + Cl (aq)
The concentration of KCl is 0.10 mol/L, we have to calculate the concentration of K^{+} and Cl^{} ion.
[K^{+}] = 1/1 x (0.10 mol/ L ) = 0.10 mol /L
[Cl^{}] = 1/1 x (0.10 mol/L) = 0.10 mol/L
Find the concentration of ions formed in the solution of 0.010 mol/L of aluminium sulphate.
We know that the decomposition of aluminium sulphate results the formation of 2 aluminium ions and 3 sulphate ions.
Al_{2}(SO_{4})_{3} (s) > 2Al^{3+}(aq) + SO_{4}^{2} (aq)
[Al^{3+}]= 2/1 x (0.010 mol/L) = 0.020 mol/L
[SO_{4}^{2}] = 3/1 x (0.010 mol/L) = 0.030 mol/L
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From Wikipedia
Molar concentration
Molar concentration, also called molarity, amount concentration or substance concentration, is a measure of the concentration of a solute in a solution, or of any chemical species in terms of amount of substance in a given volume. A commonly used unit for molar concentration used in
From Yahoo Answers
Question:Having major problems with this. Please help!
Answer these questions. Show your work.
16. What is the molar concentration (i.e., molarity) of sucrose (C12H22O11) if 150.0 g is dissolved in 250.0 mL of solution?
17. What is the molar concentration of methanol (CH3OH) if 125.0 mL is dissolved in enough water to make 15.0 L of solution? The density of methanol is 0.792 g/mL.
18. How many grams of NaOH are contained in 250.0 mL of a 0.400 M sodium hydroxide solution?
19. The drinking water standard for lead is 15 ppb (parts per billion). What is this in M?
20. What volume of 0.650 M sucrose must be diluted with water to prepare 250.0 mL of 0.423 M sucrose?
21. Which of the following solutions contains the largest concentration of chloride ions:
a. 100 mL of 0.30 M AlCl3
b. 50.0 mL of 0.60 M MgCl2
c. 200.0 mL of 0.40 M NaCl
Answers:16. Molar mass = 342 g/mol Moles = 150 / 342 = 0.439 M= 0.439 / 0.250 L = 1.75 M 17. Mass = 125 x 0.792 = 99 g Molar mass = 32 g/mol Moles = 99 / 32 = 3.09 M = 3.09 / 15 = 0.206 18. Moles = 0.250 x 0.400 = 0.10 Molar mass = 40 g/mol 40 x 0.1 = 4.0 g 19. 1 ppb = 1 mg / 1 m^3 = 1 mg / 1000 L 15 mg = 0.015 g Moles = 0.015 / 207.2 = 0.000072 M = 0.000072 / 1000 = 7.2 x 10^8 20. 0.25 x 0.423 = V x 0.65 V = 0.163 L = 163 mL 21. a. 0.3 x 0.1 = 0.03 moles AlCl3 => moles Cl = 0.09 b. 0.05 x 0.6 = 0.03 moles MgCl2 => moles Cl = 0.06 c. 0.2 x 0.4 = 0.08 moles NaCl => 0.08 moles Cl
Answers:16. Molar mass = 342 g/mol Moles = 150 / 342 = 0.439 M= 0.439 / 0.250 L = 1.75 M 17. Mass = 125 x 0.792 = 99 g Molar mass = 32 g/mol Moles = 99 / 32 = 3.09 M = 3.09 / 15 = 0.206 18. Moles = 0.250 x 0.400 = 0.10 Molar mass = 40 g/mol 40 x 0.1 = 4.0 g 19. 1 ppb = 1 mg / 1 m^3 = 1 mg / 1000 L 15 mg = 0.015 g Moles = 0.015 / 207.2 = 0.000072 M = 0.000072 / 1000 = 7.2 x 10^8 20. 0.25 x 0.423 = V x 0.65 V = 0.163 L = 163 mL 21. a. 0.3 x 0.1 = 0.03 moles AlCl3 => moles Cl = 0.09 b. 0.05 x 0.6 = 0.03 moles MgCl2 => moles Cl = 0.06 c. 0.2 x 0.4 = 0.08 moles NaCl => 0.08 moles Cl
Question:80.0 mL of 4.00 mol/L, H2SO4 diluted to 400.0 mL by adding water.
Please explain to me what happens as well.
Thank you! How is a 1.0 mol/L solution of hydrochloric acid different from a 1.0 mol/solution of acetic acid?
Answers:For the first question, the answer is relatively simple to find. Imagine that instead of mol/L of solution, you actually have milimoles/ml. The numbers are the same, only the units are different. Then, multiply that by the number of ml of solution, like so: 80.0ml*4.00mmol/ml=34.0mmol and you have the total number of mmol H2SO4 in solution. Then, divide by the new total volume of water in the solution (480.ml) to get the final molarity. For the second question, the difference is that in the acetic acid solution, there are fewer dissolved H+ ions. HCl is a strong acid, which means that in solution it dissociates completely into its component parts, H+ and CL. Acetic acid is not a strong acid, and though it does dissolve in water, it does not fall apart as readily, thus meaning that the concentration of H+ will be lower, and the PH will be higher, for a 1.0M solution of HCL than for a 1.0M solution of HC2H3O2.
Answers:For the first question, the answer is relatively simple to find. Imagine that instead of mol/L of solution, you actually have milimoles/ml. The numbers are the same, only the units are different. Then, multiply that by the number of ml of solution, like so: 80.0ml*4.00mmol/ml=34.0mmol and you have the total number of mmol H2SO4 in solution. Then, divide by the new total volume of water in the solution (480.ml) to get the final molarity. For the second question, the difference is that in the acetic acid solution, there are fewer dissolved H+ ions. HCl is a strong acid, which means that in solution it dissociates completely into its component parts, H+ and CL. Acetic acid is not a strong acid, and though it does dissolve in water, it does not fall apart as readily, thus meaning that the concentration of H+ will be lower, and the PH will be higher, for a 1.0M solution of HCL than for a 1.0M solution of HC2H3O2.
Question:I need this for homework and I can't find the answer anywhere :(
Answers:1 X 10^8 mol / L
Answers:1 X 10^8 mol / L
Question:You have 350g of Aluminum Bromide (AlBr3) with a molecular weight of 266.6938 g/mol. It is dissolved in water and the total volume of the solution is brought to 1.00L. Please explain.... Thanks so much
Answers:find moles: 350g of (AlBr3) divided by 266.69 g/mol = 1.312 moles AlBr3 since it was dissolved in 1 liter. you have 1.312 Molar AlBr3 since 1 mole of AlBr3 has 1 mole Al 1.312 Molar AlBr3 releases 1.312 Molar Al +3 ions since 1 mole of AlBr3 has 3 mole Br 1.312 Molar AlBr3 releases 3 times as much = 3.397 Molar Br 1 ions 1.312 Molar Al +3 ions + 3.397 Molar Br 1 ions = 5.249 Molar in ions total your answer, rounded to 3 sig figs is 5.25 Molar in ions
Answers:find moles: 350g of (AlBr3) divided by 266.69 g/mol = 1.312 moles AlBr3 since it was dissolved in 1 liter. you have 1.312 Molar AlBr3 since 1 mole of AlBr3 has 1 mole Al 1.312 Molar AlBr3 releases 1.312 Molar Al +3 ions since 1 mole of AlBr3 has 3 mole Br 1.312 Molar AlBr3 releases 3 times as much = 3.397 Molar Br 1 ions 1.312 Molar Al +3 ions + 3.397 Molar Br 1 ions = 5.249 Molar in ions total your answer, rounded to 3 sig figs is 5.25 Molar in ions
From Youtube
How To Calculate Molar Concentration or Molarity :A video produced for "The Smarticle Particles" class blog.
Concentrations :Finding the concentration of a solution in a variety of different units including mass %, volume %, molarity, and molality.