normality

2.4.5 Normality At the end of this section students will be able to " define the terms equivalent weight, number of equivalents and normality; " calculate normality of a solution from a given information. The normality of a solution is the gram equivalent weight of a solute per liter of solution. One gram equivalent of a solute per liter of solution is expressed as 1 normal. It is indicated using the symbol N, eq/L, or meq/L (= 0.001 N) for units of concentration. N = number of gram equivalents of solute volume of solution in litre Eq 2.14 Where number of gram equivalents of solute = mass of solute equivalent weight Eq 2.15 but equivalent weight= molar mass of solute z Eq 2.16 Substituting eq 2.16 in eq 2.15 gives: 89 Solutions number of gram equivalents of solute = mass of solute x z molar mass eq 2.17 Substituting eq 2.17 in eq 2.14 gives: N = mass of solute x Z molar mass x volume of solution in litre Eq 2.18 Where ‘z’ is the number of transferable H+ or OH- ions in acid-base reaction, positive or negative charges carried by the cations or anions in precipitation reaction, or electrons in redox reactions. Thus, z = 1 for HCl, 2 H2 SO4 (because 1 H2 SO4 releases 2 H+ ions in aqueous solution), 3 for H3 PO4 , 1 for 1 NaOH, 2 for Ca(OH)2 , 2 for Na2 CO3 , and 6 for Al2 (SO4 ) 3 (because Al2 (SO4 ) 3 dissociates to give 2Al3+ = 6 positive charges). Example 2.11 Calculate the number of gram equivalent or simply equivalents in each of the following (a) 18.25 g HCl (b) 20 g NaOH (c) 98 g H2 SO4 Given Required Solution (a) Mass solute (HCl) = 18.25 g Molar mass (HCl) = 36.5 g/mol eq The reaction: HCl → H+ + ClHCl produces one H+; thus z =1. number of gram equivalents of solute =(8.25/36.5) x 1= 0.5 eq (b) Mass solute (NaOH) = 20 g Molar mass (NaOH) = 40 g/mol eq The reaction: NaOH → Na+ + OHNaOH produces one OH- ; z =1. number of gram equivalents of solute =(20/40) x 1 = 0.5 eq (c) Mass solute (H2 SO4 ) = 98 g Molar mass = 98 g/mol eq The reaction: H2 SO4 → 2H+ + SO4 2- H2 SO4 releases two H+ ions in this case; z =2. number of gram equivalents of solute =(98/98) x 2 = 2 eq 90 Chemistry Grade 10 Example 2.12 (i) Calculate the number of equivalents present in 0.50 mol H3 PO4 if the acid is a. completely neutralized to give PO4 3- b. converted to H2 PO4 - c. converted to HPO4 2- (ii) Calculate the Normality of each of the solutions described in Q(i) a to c if the solution volume was 100 mL each. (iii) What is the mass of Na2 CO3 required to prepare 100 mL of 0.1 N Na2 CO3 ? (Given: Molar mass Na2 CO3 = 106 g/mol). Given Required Solution (a) Mole H3 PO4 = 0.5 number of equivalents The appropriate equation is H3 PO4 → 3H+ + PO4 3– H3 PO4 releases 3 H+ ions in this case; z = 3 equivalents = (mass of solute)/(molar mass) x z equivalents = n x z= 0.5 x 3 =1.5 eq Where, n = (mass solute)/(molar mass solute) =moles of solute (b) Mole H3 PO4 = 0.5 eq The appropriate equation is H3 PO4 → H+ + H2 PO4 1– H3 PO4 releases 1 H+ ion in this case; z =1 equivalents = (mass of solute)/(molar mass) x z equivalents = n x z=0.5 x 1=0.5 eq (c) Mole H3 PO4 = 0.5 eq The appropriate equation is H3 PO4 → 2H+ + H2 PO4 2 H3 PO4 releases 2 H+ ion in this case; z =2 equivalents=(mass of solute)/(molar mass) x z equivalents=n x z=0.5 x 2=1 eq Normality is the only concentration unit that is reaction dependent. For example, in a multistep dissociation of diprotic (e.g. H2 SO4 ) and polyprotic acids (e.g. H3 PO4 ), the value of ‘z’ depends on the actual number of H+ released; not just the one in the formula. 91 Solutions ii) Calculate the Normality of each of the solutions described in Q(i) a to c if the solution volume was 100 mL each. Given Required Solution (a) Mole H3 PO4 = 0.5 V = 100 mL = 0.1 L Normality a) N=Eq/L= (1.5 Eq)/(0.1 L)=15 Eq/L=15 N b) N=Eq/L= (0.5 Eq)/(0.1 L)=5 Eq/L=5 N c) N=Eq/L= (1 Eq)/(0.1 L)=15 Eq/L=10 N (iii) What is the mass of Na2 CO3 required to prepare 100 mL of 0.1 N Na2 CO3 ? (Given: Molar mass Na2 CO3 = 106 g/mol) Given Required Solution N = 0.1 eq/L V = 100 mL = 0.1 L Molar mass Na2 CO3 = 106 g/mol Mass? Na2 CO3 release 2Na+; z = 2 Thus N = mass of solute x Z molar mass x volume of solution in litre ⇒ Mass = [0.1 eq/Lx106 g/mol x 0.1 L]/2 = 0.53 g The easiest way to find normality is from molarity. All you need to know is the value of ‘z’. N = M.z; where N is normality and M is molarity Example 2.13 Calculate the normality of 0.5 M H2 SO4 solution (assume complete dissociation). Given Required Solution M = 0.5 N N = Mz = 0.5 x 2 = 1 N 1. How many equivalents of solute is contained in1 L of 2 N solution? 2. Calculate the mass of Al2 (SO4 ) 3 in 250 mL of solution if the concentration is 0.48 N. 3. Calculate the molarity and normality of a solution that contains 16.2 g of the salt Fe2 (SO4 ) 3 in 200 mL of solution. 4. Calculate the normality of a. 0.1381 M NaOH b. 0.0521 M H3 PO4 assuming complete dissociation. Exercise 2.10 92 Chemistry Grade 10 2.4.6 Conversion of Concentration Units At the end of this section students will be able to inter convert various concentration expressions. The interconversion of concentration units is very simple. Because they are intrinsic units. That means, any amount of liquid taken from a 5% (v/v) ethanol solution, for instance, always contains ethanol to solution in ratio of 5 is to100 regardless of the volume taken. Therefore, 5% (v/v) read as 5 mL of ethanol per 100 mL of solution. 5 ppm glucose means 5 g of glucose per 106 g of solution. Similarly, 5 ppb of glucose means 5 g of glucose per 109 g of solution. If percent is given in (v/v), convert the mL of solute into moles by converting volume to mass by multiplying with density followed by calculating moles using the formula (mole = mass/molar mass). And convert mL of solution into liter. Example 2.14 Converting Mole Fractions to Molality The mole fractions of glucose, C6 H12O6 , and water are 0.150 and 0.850, respectively. What is the molality of glucose in the solution? Given Required Solution Xsolute(glucose) = 0.15 Xsolvent(water) = 0.85 Molality = m Molality = m = number of moles of solute mass of solvent in kg Let’s consider 0.15 moles of glucose and 0.85 moles of water which is consistent with the given mole fractions. Thus, to calculate the mass of solvent in kg we convert the mole solvent into mass solvent as follows: MassH2 O = mole x molar mass = 0.85 mole x18 g/mole =15.3 g = 0.0153 kg Substituting the values in the above equation gives: Molality = m = 0.15 mole =9.8 m 0.0153 kg 93 Solutions Example 2.15 1. An aqueous solution is 0.273 m KCl. What is the molar concentration of potassium chloride, KCl? The density of the solution is 1.011 x 103 g/L. Given Required Solution Molality = 0.273 m Density solution = 1.011 x 103 g/L solute = KCl Molar mass KCl = 74.6 g/mole Molarity = M Molarity = M = number of moles of solute Volume of solution in litre but molality = 0.273 m means 0.273 mole of KCl per 1 kg (=1000 g) of solution. Now, we know number of moles of solute. To determine volume of solution in liter we proceed as follows: Vsolution = masssolution/dsolution = (masssolute + masssolvent)/dsolution = (0.273 mole x 74.6 g/mole) + (1000g)/ (1.011 x 103 g/L) =1.01 L Substituting the values in the above equation gives: Molarity = M = 0.273 mole =0.270 M 1.01 L Example 2.16: Converting (w/w) percent to Molarity Calculate the molarity of 36.5%(w/w) HCl solution (density of solution = 1.20 g/ mL; molar mass HCl = 36.5 g/moL) Given Required Solution %(w/w) = 36.5 Density solution = 1.20 g/mL Molar mass (HCl) = 36.5 g/mole Convert % into Molarity Molarity = M = number of moles of solute Volume of solution in litre but %(w/w) = 36.5% means 36.5 g of HCl in 100 g of solution. Thus, number of moles of HCl is calculated as follows: Number of moles of HCl = mass/molar mass = 36.5 g/36.5 g.mol-1=1 mol volume of solution = msol/dsoln = 100 g/1.2 = 83.3 mL = 0.0833 L Substituting the values in the above equation gives: Molarity = M = 1 mole =12.0 M 0.0833 L 97 Solutions And the molarity of solution II is M2 = Moles2 /V2 Rearranging the equations to find moles we obtain Moles1 =M1 V1 And Moles2 =M2 V2 As you can see in Figure 2.16, adding more solvent changed the volume of the solution. Doing so we also changed its concentration because the number of particles (number of moles) per unit volume (per unit litre or per unit mL) is changed. However, the number of particles (number of moles) of solute did not change. So, Moles1 = Moles2 M1 V1 = M2 V2 This formula is known as dilution formula. It is used to prepare dilute solutions. For instance, to prepare a dilute solution having a volume V2 and concentration M2 using a more concentrated solution having concentration M1 , we withdraw the volume V1 from the more concentrated (stock) solution having the concentration M1 and add more water to raise the volume from V1 to V2 . We often need to calculate V1 using the formula: V1 =M2 V2 /M1 Caution! When preparing dilutions of concentrated acids such as sulfuric acid, the directions usually call for adding the acid slowly to water with frequent stirring. If water were added to acid, the water would quickly heat and splatter, causing harm to the person making the solution. Example 2.18 Describe how you would prepare 2.50 x 102 mL of a 2.25 M H2 SO4 solution, starting with a 7.41 M stock solution of H2 SO4 . Given Required Solution M1 = 7.41 M V1 = ? M2 =2.25 M V2 = 250 mL V1 M1 V1 = M2 V2 V1 = M2 V2 M1 V1 =(2.25 M x 250 mL)/(7.41 M) =75.9 mL Thus, we must withdraw 75.9 mL from the 7.41 M H2 SO4 solution, transfer it to a 250 mL volumetric flask half-filled with distlled water, and dilute it with sufficient amount of water to the mark. 98 Chemistry Grade 10 1. How would you prepare 2.00 x 102 mL of a 0.866 M KOH solution, starting with a 5.07 M stock solution? 2. What is the difference between dilute and unsaturated? Or concentrated and saturated? 3. Prepare a 100 mL of 0.1 M sucrose from 1M sucrose solution by dilution. Exercise 2.12 Outline the preparation of 100 mL of (a) hypotonic (0.1 M) and (b) hypertonic (0.25 M) aqueous NaCl solution and (c) 0.1 M H2 SO4 using a stock solution of 98% (w/w) H2 SO4 . Density = 1.84 g/ml. (Hint: First calculate the molarity of the 98%(w/w) H2 SO4 solution as shown in example 2.16. Then use the dilution formula). Show all the necessary steps and let your teacher check it! Exercise 2.13 2.6 Solution Stoichiometry At the end of this section students will be able to " use stoichiometrically equivalent molar ratios to calculate amounts of reactants and products in a reaction of pure and dissolved substance. In Unit 3 you studied about stoichiometric calculations in terms of the mole method, which treats the coefficients in a balanced equation as the number of moles of reactants and products. Solving stoichiometry problems for reactions in solution requires the additional step of converting the volume of reactant or product in solution to amount (mole) using the relationship MV = number of moles of solute. Generally, we follow these steps: 1. Balance the equation. Note that the coefficients in chemical equation indicate number of moles. 2. Find the amount (mol) of one substance from the volume and molarity. 3. Relate it to the stoichiometrically equivalent amount of another substance. 4. Convert to the desired units. 99 Solutions Mole-Mole Calculation Example 2.19 What is the concentration of sodium hydroxide required to react completely with equal volume of 0.104 M sulfuric acid? Given Required Solution VNaOH= VH2 SO4 MH2 SO4 = 0.104 M MNaOH = ? Equation: H2 SO4 + 2NaOH → Na2 SO4 + 2H2 O Stoichiometry 1 mole 2 mole Equation H2 SO4 + 2NaOH → Na2 SO4 + 2H2 O Given n H 2 S O 4= 0 . 1 0 4 mol/L x VH2 SO4 nNaOH=MNaOH x VNaOH Stoichiometry:= nNaOH/nH2SO4 =2/1 = MNaOHVNaOH/MH2SO4 VH2SO4 = MNaOHVNaOH/0.104MVH2SO4 Since volume is the same, it cancels out and the equation becomes: 2/1= MNaOH/0.104 M ⇒ MNaOH= 0.104 M x 2 = 0.208 M What is the concentration of sodium hydroxide that is required to react completely with equal volume of 0.104 M hydrochloric acid? Exercise 2.14 Mole - Mass Example 2.20 What mass of solid magnesium hydroxide can be produced if 45 mL of a 0.63 M Mg(NO3 ) 2 solution reacts completely with excess NaOH? Given Required Solution VMg(NO3) 2 = 45 mL MMg(NO3) 2 = 0.63 M Note that M and V are given means number of moles (n) is known: n=MV VAgNO3 = ? Stoichiometry 1 mole 58.321 g Equation 2NaOH(aq) +1Mg(NO3 ) 2 (aq) → Mg(OH)2 (s) +2NaNO3 (aq) Given 0.63 M, 45 mL Mass =? Stoichiometry:=nMg(NO3) 2 /massMg(OH)2 =1 mole/58.321 g =0.63 mol/L x 0.045 L/massMg(OH)2 ⇒massMg(OH)2 = 0.63 mol/L x 0.045L x 58.321 g/1 mole =1.65g 100 Chemistry Grade 10 Mass - Volume Example 2.21 What volume of 0.0995 M Al(NO3 ) 3 will react with 3.66 g of Ag according to the following chemical equation? Given Required Solution wAg=3.66 g MAl(NO3) 3 = 0.0995 M VAl(NO3) 3 =? Stoichiometry 3x107.97 g 1 mole Equation 3Ag(s) + Al(NO3 ) 3 (aq) → 3 AgNO3 (aq) + Al(s) Experimental 3.66 g n Al(NO3 ) 3 = MAl(NO3) 3 x VAl(NO3) 3 Stoichiometry: =wAg/nAl(NO3) 3 =3x107.97 gAg/1moleAl(NO3) 3 = 3.66g/MAl(NO3) 3 VAl(NO3) 3 =3.66g/0.0995MxVAl(NO3) 3 ⇒ 3x107.97g/1mole=3.66g/(0.0995MxVAl(NO3) 3 ); where MV = mole VAl(NO3) 3 = 3.66 g x1 mole/3x107.97 g x0.0995 mol/L=0.114 L What volume of 0.512 M NaOH will react with 17.9 g of H2 C2 O4 (s) according to the following chemical equation? H2 C2 O4 (s) + 2NaOH(aq) → Na2 C2 O4 (aq) + 2H2 O(l) Exercise 2.15 Mole - Volume Calculation Example 2.22 What volume of a 0.35 M AgNO3 is required to react completely with 55mL of a 0.24 M NaCl solution? Given Required Solution VNaCl=55 mL MNaCl=0.24 M Note that M and V are given means number of moles (n) is known: n=MV VAgNO3 =? Stoichiometry 1 mole 1 mole Equation NaCl(aq) + AgNO3 (aq) → NaNO3 (aq) + AgCl(s) Experimental 0.24 M, 55 mL 0.35 M, V=? Stoichiometry: = nNaCl/nAgNO3 =1/1 = (0.24 mol/L x0.055L)/(0.35 mol/L x VAgNO3 ) =1 VAgNO3 = (0.24 mol/L x0.055L)/(0.35 mol/L) = 0.0377 L=37.7 mL 101 Solutions Mole - Number of Particles Example 2.23 1. How many HCl particles are there in 0.1 mole of the HCl solution? Strategy: 1 mole HCl contains Avogadro’s number of HCl particles, then 0.1 mole contains how many? Answer: 0.6022 x 1023 particles = 6.022 x 1022 HCl particles What volume of 0.5 M glucose (C6 H12O6 ) contains 3.01 × 1022 molecules of glucose? Exercise 2.16 2.7 Describing Reactions in Solution At the end of this section students will be able to " explain the relationship between reacting ions, spectator ions, precipitation and solubility; " write net ionic equations. Molecular Equation A molecular equation is a chemical equation in which the reactants and the products are written as if they were molecular substances, even though they may actually exist in solution as ions. The state (solid, liquid, gas) of each substance is indicated in parentheses after formula. (aq) is used to indicate the that molecule is dissolved in water. Consider the reaction between silver nitrate, AgNO3 , and sodium chloride, NaCl, in aqueous solution to give solid silver chloride, AgCl, and aqueous sodium nitrate. The equation for this reaction may be written as: AgNO3 (aq) + NaCl(aq) → AgCl(s) + NaNO3 (aq) This is just a molecular equation. Note from the solubility rules (Table 2.1) that all common nitrates and chlorides of Group 1A are soluble. That is why (aq) is used after these formulae. The solubility rule also taught us that AgCl is insoluble in water. That is why (s) to indicate that it is precipitated out from the solution to the bottom of the reaction flask/test tube as a solid. 102 Chemistry Grade 10 Ionic Equation Although a molecular equation is useful in describing the actual reactants and products, it does not give any information about what is happening at the level of ions. Since this kind of information is very useful, you often need to write the molecular equation as an ionic equation. Consider the reaction of silver nitrate and sodium chloride. Both are soluble ionic substances and are strong electrolytes. When they dissolve in water, they go into solution as ions giving on the reactant side of the reaction: Ag+(aq) + NO3 (aq) + Na+(aq) + Cl (aq) On the product side of the equation, AgCl(s) is an ionic compound that doesn’t dissolve in water, but the ions are held together in particular sites in the crystalline solid. We leave the formula as AgCl(s) to convey this information in the equation. On the other hand, NaNO3 is a soluble ionic compound and is a strong electrolyte. Also it dissolves in aqueous solution to give freely moving ions. Therefore, we can write it as Na+(aq) + NO3 – (aq) The complete equation is Ag+(aq) + NO3 (aq) + Na+(aq) + Cl (aq) → AgCl(s) + Na+(aq) + NO3 (aq), The ions appearing on both sides of the equation (Na+ and NO3 – ) are called spectator ions, as they do not take part in the reaction and they can be canceled on both sides to express the essential reaction that occurs. Ag+(aq) + NO3 (aq) + Na+(aq) + Cl_ (aq) → AgCl(s) + Na+(aq) + NO3 (aq) The resulting equation is: Ag+ (aq) + Cl (aq) → AgCl(s) This net ionic equation, is without spectator ions and the reaction that actually occurs at an ionic level is between silver ions and chloride ions which forms solid silver chloride. Steps for writing a net ionic equation 1. Write the balanced molecular equation 2. Write the balanced complete ionic equation 3. Cancel out the spectator ions. What remains is the net ionic equation.