1) Expression of concentration molar concentration percent concentration conversion of units Chemical calculations Vladimíra Kvasnicová 2) Osmotic pressure, osmolarity 3) Dilution of solutions 4) Calculation of pH strong and weak acids and bases buffers Important terms Important terms solute = a substance dissolved in a solvent in forming a solution density (ρ) = the mass of a substance per unit of volume (kg.m-3 or g.cm-3) ρ = m/V solvent = a liquid that dissolves another substance or substances to form a solution mass solution = a homogeneous mixture of a liquid (the solvent) with a gas or solid (the solute) concentration = the quantity of dissolved substance per unit quantity of solution or solvent m = n x MW (in grams) amount of substance (n) = a measure of the number of entities present in a substance (in moles) Avogadro constant (NA) = the number of entities in one mole of a substance (NA = 6.022x1023) molar weight (MW) = mass of one mole of a substance in grams (in g/mol) Important terms relative molecular mass (Mr) = the ratio of the average mass per molecule of the naturally occurring form of an element or compound to 1/12 of the mass of 12C atom molar concentration (= molarity) M = mol/1000 mL 1M solution ⇒ 1 mol of a solute is found in 1000 mL (= 1L) of the solution 0,5M solution ⇒ 0,5 mol of a solute is found in 1000 mL (= 1L) of the sol. Mr = sum of relative atomic masses (Ar) of all atoms that comprise a molecule percent concentration MW (grams / mol) = Mr 1% solution dilution = process of preparing less concentrated solutions from a solution of greater concentration ⇒ 1 g of a solute is found in 100 g of the solution Expression of concentration Molarity (c) (mol x l-1 = mol x dm-3 = M ) = number of moles per liter of a solution c = n / V % = g/100g 0,5% solution ⇒ 0,5 g of a solute is found in 100 g of the solution 1M NaOH MW = 40g /mol => 1M solution of NaOH = 40g of NaOH / 1L of solution 0,1M solution of NaOH = 4g of NaOH / 1L of solution Preparation of 500 mL of 0,1M NaOH: 0,1M solution of NaOH = 4g of NaOH / 1 L of solution number of moles / 1000 mL of solution DIRRECT PROPORTIONALITY 2g of NaOH / 0.5 L of solution ! DIRRECT PROPORTIONALITY ! Problems 1) 17,4g NaCl / 300mL, MW = 58g/mol, C = ? [1M] 2) Solution of glycine, C = 3mM, V = 100ml. ? mg of glycine are found in the solution? [22,5mg] 3) Solution of CaCl2, C = 0,1M. Calculate volume of the sol. containing 4 mmol of Cl-. [20ml] Normality (N) = concentration in terms of equivalent weights of substance (reflect the number of combining or replaceable units). It is not in common use! 1M 1M 1M 1M 1M HCl = H2SO4 = H3PO4 = CaCl2 = CaSO4 = 1N 2N 3N 2N 2N HCl H2SO4 H3PO4 CaCl2 CaSO4 Molality (mol.kg –1) = concentration in moles of substance per 1 kg of solvent Osmolality ( mol.kg –1 or osmol.kg -1) = concentration of osmotic effective particles (i.e. particles which share in osmotic pressure of solution) • it is the same (for nonelectrolytes) or higher (for electrolytes: they dissociate to ions) as molality of the same solution Osmolarity (osmoles / L) = osmolality expressed in moles or osmoles per liter the semipermeable membrane separates two solutions of different concentrations the passage of a solvent through a semipermeable membrane is called osmosis http://www.phschool.com/webcodes10/index.cfm?error=1&errortype=default_global& errortcode= Problems Describe dissociation of the salts: osmotic pressure osmolarity = molarity of all particles dissolved in a solution (= osmotic active particles) KNO3 → K+ + NO3- Σ 2 ions K2CO3 → 2 K+ + CO32- Σ 3 ions Na3PO4 → 3 Na+ + PO43- Σ 4 ions Na2HPO4 → 2 Na+ + HPO42- Σ 3 ions NaH2PO4 → Na+ + H2PO4- Σ 2 ions NH4HCO3 → NH4+ + HCO3- Σ 2 ions What is the osmolarity of the 1M solutions? http://www.biologycorner.com/resources/osmosis.jpg http://www.mhhe.com/biosci/esp/2001_gbio/folder_structure/ce/m3/s3/assets/image s/cem3s3_1.jpg http://campus.queens.edu/faculty/jannr/cells/cell%20pics/osmosisMicrographs.jpg Osmotic pressure (Pa) π = i x c x R x T i = 1 (for nonelectrolytes) i = number of osmotic effective particles (for strong electrolytes) isotonic solutions = solutions with the same value of the osmotic pressure (c.g. blood plasma x saline ) Oncotic pressure = osmotic pressure of coloidal solutions, e.g. proteins Percent concentrations • generally expressed as parts of solute per 100 parts of total solution (percent or „per one hundred“) • three basic forms: a) weight per unit weight (W/W) g/g of solution 10% NaOH → 10g of NaOH + 90g of H2O = 100g of sol. 10% KCl → 10g of KCl/100g of solution Problems 4) ? osmolarity of 0,15mol/L solution of : a) NaCl b) MgCl2 c) Na2HPO4 d) glucose [0,30 [0,45 [0,45 [0,15 M] M] M] M] 5) Saline is 150 mM solution of NaCl. Which solutions are isotonic with saline? [= 150 mM = 300 mosmol/l ] a) b) c) d) 300 mM glucose 50 mM CaCl2 300 mM KCl 0,15 M NaH2PO4 [300] [150] [600] [300] b) volume per unit volume (V/V) ml/100ml of sol. 5% HCl = 5ml of HCl / 100ml of sol. c) weight per unit volume (W/V) g/100 ml (g/dl; mg/dl; µg/dl; g % ) • the most frequently used expression in medicine 20% KOH = 20g of KOH / 100 ml of sol. Problems 6) 600g 5% NaCl, ? mass of NaCl, mass of H2O [30g NaCl + 570g H2O] 7) 250g 8% Na2CO3, ? mass of Na2CO3 (purity 96%) [20,83g {96%}] 8) Normal saline solution is 150 mM. What is its percent concentration? [ 0,9%] Problems 9) 14g KOH / 100ml 10) C(HNO3) = 5,62M; ρ = 1,18g/cm3 (density), MW = 63g/mol, ? % [ 30% ] 11) 10% HCl; ρ = 1,047g/cm3, MW = 36,5 g/mol ? C(HCl) [ 2,87M ] Problems • Calculate the molar concentration of 30% HNO3, if its density ρ = 1,18 g/cm3, MW = 63 g/mol. (5,62 M) • Calculate the percent concentration of 2,87M HCl if ρ = 1,047 g/cm3, MW = 36,5 g/mol. (10%) • Calculate the molar concentration of a solution containing 14 g of KOH in 100 mL (MW = 56,1 g/mol). Use the simplification: 1mL of this solution = 1g. (2,5 M) • Calculate the molarity of 70% HClO4 (ρ = 1,67g/cm3, MW = 100,5 g/mol). (11,63 M) • Calculate the percent concentration of 11,63 M HClO4 (ρ = 1,67g/cm3, MW = 100,5 g/mol). (70%) MW = 56,1g/mol; C = ? [ 2,5M ] Problems • Your task is to prepare 2 L of 0,1 M HCl. A bottle of 35% HCl is labeled: 1 L = 1,18 kg, molar weight of HCl is 36,5 g/mol. How many (17,68 mL) millilitres of 35% do you need? • Your task is to prepare 250 mL 0,1 M of ammonia. A bottle of 25% ammonia is labeled: 1 L = 0,91 kg, molar weight of ammonia is 17,0 g/mol. How many millilitres of 25% ammonia do you need? (1,87 mL) • Your task is to prepare 1 L of 1 M acetic acid. A bottle of 99% acetic acid is labeled: 1 L = 1,05 kg, molar weight of acetic acid is 60,0 g/mol. How many millilitres of this 99% acid do you need? (57,7 mL) • Your task is to prepare 0,5 L of 1 M HClO4. A bottle of 70% HClO4 is labeled: 1 L = 1,67 kg, molar weight of HClO4 is 100,5 g/mol. How many millilitres of 70% acid do you need? (42,99 mL) A concentration (physilological ranges) of substances analyzed in blood can be expressed by various units. Add values in alternative units into the table below (molarity is used in ČR ↔ mass concentration is used in foreign literature) Results: MW (g/mol) mass concentration (in mg/dL or µg/dL) molar concentration (in mmol/l or µmol/l) analyte MW (g/mol) mass concentration (in mg/dL or µg/dL) molar concentration (in mmol/l or µmol/l) analyte glucose 180 70 - 106 mg/dL .................... mmol/L glucose 180 70 - 106 mg/dL 3,9 - 5,9 mmol/L lactic acid 90 .................... mg/dL 0,5 - 2,2 mmol/L lactic acid 90 4,5 - 20 mg/dL 0,5 - 2,2 mmol/L 387 up to 200 mg/dL up to ................... mmol/L cholesterol 387 up to 200 mg/dL up to 5,2 mmol/L cholesterol 17 .................... µg/dL 16 - 53 µmol/L ammonia 17 27 - 90 µg/dL 16 - 53 µmol/L ammonia 585 0,1 - 1,2 mg/dL .................... µmol/L bilirubin 585 0,1 - 1,2 mg/dL 2 - 21 µmol/L bilirubin uric acid 168 3,6 - 8,2 mg/dL 214 - 488 µmol/L uric acid 168 ................... mg/dL 214 - 488 µmol/L creatinine 113 0,84 - 1,25 mg/dL 74 - 110 µmol/L creatinine 113 0,84 - 1,25 mg/dL .................... µmol/L urea 60 17 - 43 mg/dL 2,8 - 7,2 mmol/L urea 60 .................... mg/dL 2,8 - 7,2 mmol/L