Concentration is defined as amount of solute per amount of solution or solvent. The "amounts" you chose is what gives each concentration term it's own uniqueness. Show
More on "ppm" ... Remember that ppm is parts-per-million and is like percentage except that it is based on 1 million parts (not 100). So by mass, \[\rm ppm = {mass\;of\;solute\over mass\;of\;solution} \times 10^6 \] In dilute aqueous solutions, the density is the same as water (1 g/mL) which means that \({\rm 1\;ppm \approx 1\; mg/L}\). So anytime you see "ppm" change it to "mg/L" for all our calculations. Then you can easily convert that to mol/L to get concentration in molarity. For the rest of this page, concentration is assumed to be in molarity although most all concentration terms will work out the same way in that they are all intensive properties.Molarity
Dilutions - adding more solvent (ie. water) to an already existing solution\[M_1\cdot V_1 = M_2\cdot V_2\] ANY concentration and volume unit will work here as long as the same units are on BOTH sides of the equation. Titrations (1:1 ratios)For simple 1:1 ratios of acid and base: \({\rm A}+{\rm B}\rightarrow {\rm water\;+\;salt}\) \[M_{\rm acid}\cdot V_{\rm acid} = M_{\rm base}\cdot V_{\rm base}\] NOTE: This equation only works for monoprotic acids (one proton in the formula) and monobases (one hydroxide in the formula). Why? Because then the moles of acid will equal the moles of base, a nice 1:1 ratio. If you have ANY other ratio (not 1:1) you must use the following formula: Titrations (any ratio)For any acid/base ratio (\(a:b\)) of acid and base: \(a{\rm A}+b{\rm B}\rightarrow {\rm water\;+\;salt}\) \[{M_{\rm acid}\cdot V_{\rm acid}\over a} = {M_{\rm base}\cdot V_{\rm base}\over b}\] Note that \(a\) and \(b\) are the coefficients for the acid and base in the balanced chemical equation. For a given volume of any solution, you can always calculate the number of moles of solute... \[M_{\rm solute}\cdot V_{\rm solution} = {\rm moles\;of\;solute} = n_{\rm solute}\] To get moles you must use molarity for the concentration and liters as the volume. If you use milliliters (mL) for volume you will get millimoles (mmol) for the amount. Be careful and note which type unit you are using. Dr. McCord will often use mL with molarity and get mmol for \(n\). Always make sure you know what units you are using. If you pour together multiple solutions of the same solute, you must calculate the number of moles in each separate solution first, then add all the moles together, and divide by the total combined volume. As long as you quantify all the mole sources (whether solutions, solids, or whatever) and the total accumulated volume, you can easily calculate the concentration. Mixing Together 3 solutions of the same soluteMix solutions A, B, and C (all different concentrations and volumes, but all the same solutes) \[M_{\rm mixture} = {M_{\rm A}V_{\rm A} + M_{\rm B}V_{\rm B} + M_{\rm C}V_{\rm C} \over V_{\rm A}+V_{\rm B}+V_{\rm C}}\] Note that the numerator here is just a total (sum) of all the moles of solute and the denominator is a total (sum) of all the volumes. Acid/Base FormulasBe sure and know how to get any one of these values from any one of the others...
the value for \(K_{\rm w}\) is only valid at 25 C More Tutorials Join AUS-e-TUTE Contact Us Want chemistry games, drills, tests and more? You need to Join AUS-e-TUTE! Key Concepts ⚛ Concentration of a solution refers to the amount of solute dissolved in a given solvent to make a solution.
⚛ Concentration of a solution can be given in moles of solute per litre of solution (mol L-1 or mol/L or M), or, in moles of solute per cubic decimetre of solution (mol dm-3 or mol/dm3)(1)
⚛ Molarity is the term used to describe a concentration given in moles per litre. Alternative names for molarity are · amount of substance concentration (IUPAC preferred term) · amount concentration · molar concentration ⚛ Amount of substance concentration, molarity, has the units mol L-1 (or mol/L or M) or the equivalent SI units of mol dm-3 (mol/dm3) ⚛ Amount of substance concentration, molarity, the concentration of a solution in mol/L, mol L-1, mol dm-3 or mol/dm3, is given the symbol c (sometimes M). For a 0.01 mol L-1 HCl(aq) solution we can write : (i) [HCl(aq)] = 0.01 mol L-1 = 0.01 mol dm-3 = 0.01 M (ii) c(HCl(aq)) = 0.01 mol L-1 = 0.01 mol dm-3 = 0.01 M ⚛ Mathematical equation (formula or expression) to calculate the molarity of a solution (concentration in mol L-1) is c = n ÷ V c = concentration of solution in mol L-1 (mol/L or M) ⚛ This equation (formula or expression) can be re-arranged to find: (i) moles of solute given molarity and volume of solution: n = c × V (ii) volume of solution given moles of solute and molarity: V = n ÷ c Please do not block ads on this website. Molarity, amount of substance concentration, ConceptsAmount of substance concentration, molarity, is the term given for concentrations of solutions that are given in units of moles per litre (mol L-1 or mol/L or M) or, in SI units, moles per cubic decimetre (mol dm-3 or mol/dm3).
↪ Back to top Molarity Equation (amount of substance concentration equation)The amount of substance concentration, or the molarity of a solution, is given by the equation : c = amount concentration (molarity) in mol L-1 (or mol dm-3) Note that we often use square brackets around the formula of solvated species to indicate the molarity of a solution (the concentration of the solution in mol L-1): for example an aqueous solution of sodium chloride, NaCl(aq), with a molarity of 0.154 mol L-1 (or 0.154 mol dm-3) could be represented as c(NaCl(aq)) = 0.154 mol L-1 = 0.154 mol dm-3 [NaCl(aq)] = 0.154 mol L-1 = 0.154 mol dm-3 If you know the moles of solute in a solution, and, you know the volume of the solution, you can calculate the concentration of the solution in mol L-1 (or mol dm-3) using the mathematical equation c = n ÷ V But what if you know the concentration of the solution in mol L-1 (or mol dm-3) and the volume of the solution in L (or dm3), can you calculate how many moles of solute are present in the solution? You can even calculate the volume of the solution if you know how many moles of solute are present and its concentration in mol L-1 (or mol dm-3). In summary, to calculate (a) amount of substance concentration (molarity) in mol L-1 (or mol dm-3): c = n ÷ V (b) amount of solute in solution in mol : n = c × V (c) volume of solution in L (or dm3): V = n ÷ c ↪ Back to top Examples of Molarity Calculations with Worked SolutionsApply the following 5 steps to solve the molarity problems below:
(1) Calculating Amount of Substance Concentration, Molarity (c = n ÷ V)Question: Calculate the amount of substance concentration in mol L-1 (molarity) of an aqueous sodium chloride solution containing 0.125 moles sodium chloride in 0.50 litres of solution. Solution: Step 1: What is the question asking you to calculate? c(NaCl(aq)) = molarity (concentration in mol L-1) of solution = ? mol L-1 Step 2: What information has been given in the question? Extract the data from the question Step 3: What is the relationship between what you know and what you need to find? Write the equation: Step 4: Substitute the values into the equation for molarity and solve: [NaCl(aq)] = c(NaCl(aq)) = 0.125 mol ÷ 0.50 L = 0.25 mol L-1 (or 0.25 mol/L or 0.25 M) (Note: only 2 significant figures are justified) Step 6: Write the answer: [NaCl(aq)] = 0.25 mol L-1 (2) Calculating Amount of Solute (n = c × V)Question: Calculate the moles of copper sulfate in 250.00 mL of 0.020 mol L-1 copper sulfate solution, CuSO4(aq). Solution: Step 1: What is the question asking you to calculate? n(CuSO4) = moles of solute = ? mol Step 2: What information has been given in the question? Extract the data from the question: Step 3: What is the relationship between what you know and what you need to find? Write the equation: Step 4: Substitute the values into the molarity equation and solve: n(CuSO4) = 0.020 mol L-1 × 0.25000 L = 0.005000 mol = 0.0050 mol (Note: only 2 significant figures are justified) Step 5: Write the answer: n(CuSO4(aq)) = 0.0050 mol (3) Calculating Volume of Solution (V = n ÷ c)Question: Calculate the volume in litres of a 0.80 mol L-1 aqueous solution of potassium bromide containing 1.60 moles of potassium bromide. Solution: Step 1: What is the question asking you to calculate? V(KBr(aq)) = volume of solution in litres = ? L Step 2: What information has been given in the question? Extract the data from the question: Step 3: What is the relationship between what you know and what you need to find? Write the equation: Step 4: Substitute the values into the equation and solve: V(KBr(aq)) = 1.60 mol ÷ 0.80 mol L-1 = 2.00 L = 2.0 L (Note: only 2 significant figures are justified) Step 5: Write the answer: V(KBr(aq)) = 2.0 L ↪ Back to top Problem Solving: Amount of Substance Concentration (molarity)The Problem: Chris the Chemist has been given a 250.00 mL volumetric flask and asked to use it to make an aqueous solution of sodium chloride, NaCl, with a concentration of 0.100 mol L-1 for a corrosion experiment. The sodium chloride, NaCl, is available as an analytical reagent composed of white crystals. Determine the mass in grams of sodium chloride that Chris the Chemist will need to weigh out. Solving the Problem using the StoPGoPS model for problem solving:
↪ Back to top Sample Question: Mass Concentration A 25.00 mL aliquot of 0.0264 mol L-1 AgNO3(aq) is added to a 150.00 mL volumetric flask and distilled water is added until the meniscus sits on the mark when viewed at eye-level. ↪ Back to top
Footnotes: (1) There are many other ways to measure concentration of solutions in chemistry. Some common ones are listed in the introduction to solutions tutorial. (2) This is an important assumption because the volumetric flask has been calibrated to hold the stated volume at a given temperature. Volume is dependent on temperature. (3) This is an important assumption because if the water used to make the solution already contains dissolved sodium chloride we do not know how much sodium chloride is already present. Any dissolved sodium chloride in the water will increase the concentration of sodium chloride in the final solution. (4) This is an important assumption because if the reagent contains impurities, some of the mass we weigh out will NOT be due to NaCl and the concentration of NaCl will actually be less than we think. ↪ Back to top |