1. 1 Density Calculations In this section, you will learn how to find the molarity of solution from two pieces of information (density and percentage). Usually the calculation is simple and can be done using several procedures. Look at the examples below:

2. 2 Example What volume of concentrated HCl (FW = 36.5g/mol, 32%, density = 1.1g/mL) are required to prepare 500 mL of 2.0 M solution. Solution Always start with the density and find how many grams of solute in each mL of solution.

3. 3 Remember that only a percentage of the solution is solute. g HCl/ml = 1.1 x 0.32 g HCl / mL The problem is now simple as it requires conversion of grams HCl to mmol since the molarity is mmol per mL mmol HCl = 1.1x0.32 x10 3 mg HCl/(36.5 mg/mmol) = 9.64 mmol M = 9.64 mmol/mL = 9.64 M

4. 4 Now, we can calculate the volume required from the relation M i V i (before dilution) = M f V f (after dilution) 9.64 x VmL= 2.0 x 500mL VmL = 10.4 mL This means that 10.4 mL of the concentrated HCl should be added to distilled water and the volume should then be adjusted to 500 mL

5. 5 Example How many mL of concentrated H 2 SO 4 (FW = 98.1 g/mol, 94%, d = 1.831 g/mL) are required to prepare 1 L of 0.100 M solution? g H 2 SO 4 / mL = 1.831 x 0.94 g /mL Now find the mmol acid present mmol H 2 SO 4 = (1.831 x 0.94 x 10 3 mg) / (98.1 mg/mmol) M = mmol/mL = [(1.831 x 0.94 x 10 3 mg) / (98.1 mg/mmol)] / mL = 17.5 M

6. 6 To find the volume required to prepare the solution M i V i (before dilution) = M f V f (after dilution) 17.5 x V mL = 0.100 x 1000 mL V mL = 5.71 mL which should be added to distilled water and then adjusted to 1 L.

7. 7 An Easy Short-Cut The percentage is a fraction: (i.e. a 35% is written as 0.35) Density * percentage * 10 3 Formula Weight M =

8. 8 Analytical Versus Equilibrium Concentration When we prepare a solution by weighing a specific amount of solute and dissolve it in a specific volume of solution, we get a solution with specific concentration. This concentration is referred to as analytical concentration. However, the concentration in solution may be different from the analytical concentration, especially when partially dissociating substances are used.

9. 9 An example would be clear if we consider preparing 0.1 M acetic acid (weak acid) by dissolving 0.1 mol of the acid in 1 L solution. Now, we have an analytical concentration of acetic acid (HOAc) equals 0.1 M. But what is the actual equilibrium concentration of HOAc? We have HOAc = H + + OAc - The analytical concentration ( C HOAc ) = 0.1 M C HOAc = [HOAc] undissociated + [OAc - ] The equilibrium concentration = [HOAc] undissociated.

10. 10 For good electrolytes which are 100% dissociated in water the analytical and equilibrium concentrations can be calculated for the ions, rather than the whole species. For example, a 1.0 M CaCl 2 in water results in 0 M CaCl 2, 1.0 M Ca 2+, and 2.0 M Cl - since all calcium chloride dissociates in solution. For species x we express the analytical concentration as C x and the equilibrium concentration as [x]. All concentrations we use in calculations using equilibrium constants are equilibrium concentrations.

11. 11 Dilution Problems In many cases, a dilution step or steps are involved in analytical procedures. One should always remember that in any dilution the number of mmoles of the initial (concentrated) solution is equal to the number of mmoles of the diluted solution. This means M i V i (concentrated) = M f V f (dilute)

12. 12 Example Prepare 200 mL of 0.12 M KNO 3 solution from 0.48 M solution. M i V i (concentrated) = M f V f (dilute) 0.48 x V mL = 0.12 x 200 V mL = (0.12 x 200)/0.48 = 50 mL Therefore, 50 mL of 0.48 M KNO 3 should be diluted to 200 mL to obtain 0.12 M solution

13. 13 Example A 5.0 g Mn sample was dissolved in 100 mL water. If the percentage of Mn (At wt = 55 g/mol) in the sample is about 5%. What volume is needed to prepare 100 mL of approximately 3.0x10 -3 M solution. Solution First we find approximate mol Mn in the sample = 5.0 x (5/100) g Mn/(55 g/mol) = 4.5x10 -3 mol

14. 14 Molarity of Mn solution = (4.5x10 -3 x 10 3 mmol)/100 mL = 4.5x10 -2 M The problem can now be solved easily using the dilution relation M i V i (concentrated) = M f V f (dilute) 4.5x10 -2 x V mL = 3.0x10 -3 x 100 V mL = (3.0x10 -3 x 100)/4.5x10 -2 = 6.7 mL Therefore, about 6.7 mL of the Mn sample should be diluted to obtain an approximate concentration of 3.0x10 - 3 M solution.

15. 15 Example What volume of 0.4 M Ba(OH) 2 should be added to 50 mL of 0.30 M NaOH in order to obtain a solution that is 0.5 M in OH -. Solution We have to be able to see that the mmol OH - coming from Ba(OH) 2 and NaOH will equal the number of mmol of OH - in the final solution, which is mmol OH - from Ba(OH) 2 + mmol OH - from NaOH = mmol OH - in final solution

16. 16 The mmol OH - from Ba(OH) 2 is molarity of OH - times volume and so are other terms. Molarity of OH - from Ba(OH) 2 is 0.8 M (twice the concentration of Ba(OH) 2, and its volume is x mL. Now performing the substitution we get 0.8 * x + 0.30 * 50 = 0.5 * (x + 50) x = 33 mL

17. 17 For Solid Solutes % (w/w) = [weight solute (g)/weight sample (g)] x 100 ppt (w/w) = [weight solute (g)/weight sample (g)] x 1000 ppm (w/w) = [weight solute (g)/weight sample (g)] x 10 6 ppb (w/w) = [weight solute (g)/weight sample (g)] x 10 9 A ppm can be represented by several terms like the one above, (mg solute/kg sample), ( g solute/10 6 g sample), etc..

18. 18 If the solute is dissolved in solution we have % (w/v) = [weight solute (g)/volume sample (mL)] x 100 ppt (w/v) = [weight solute (g)/volume sample (mL)] x 1000 ppm (w/v) = [weight solute (g)/volume sample (mL)] x 10 6 ppb (w/v) = [weight solute (g)/volume sample (mL)] x 10 9 Also a ppm can be expressed as above or as (g solute/10 6 mL solution), (mg solute/L solution), or (mg/mL), etc..

19. 19 For Liquid Solutes % (v/v) = [volume solute (mL)/volume sample (mL)] x 100 ppt (v/v) = [volume solute (mL)/volume sample (mL)] x 1000 ppm (v/v) = [volume solute (mL)/volume sample (mL)] x 10 6 ppb (v/v) = [volume solute (mL)/volume sample (mL)] x 10 9 A ppm can be expressed as above or as  L/L), (mL/10 3 L), etc..

20. 20 Example A 2.6 g sample was analyzed and found to contain 3.6  g zinc. Find the concentration of zinc in ppm and ppb. A ppm is microgram solute per gram sample, therefore Ppm Zn = 3.6  g Zn/2.6 g sample = 1.4 ppm A ppb is nanogram solute/gram sample, therefore Ppb Zn = 3.6 x10 3 ng Zn/2.6 g sample = 1400 ppb