Recap – Stoichiometry Balanced chemical equation Convert mass to moles Chem 1001 Lecture 11 Recap – Stoichiometry Balanced chemical equation Convert mass to moles Check for limiting reagent Use moles and equation coefficients to obtain amount of target needed Covert moles of target to mass Calculate % yield Amount in moles, n = m / M

Reactions in Solution Many reactions occur in solution. Chem 1001 Lecture 14 Reactions in Solution Many reactions occur in solution.   Dr Toby Hudson

Chem 1001 Lecture 14 Reactions in Solution The solvent is the substance that does the dissolving. The solute is the substance that is dissolved. The solution is the mixture. Concentration refers to amount of solute dissolved in 1 L of solution.   Dr Toby Hudson

Chem 1001 Lecture 14 Reactions in Solution In chemistry we use: no moles solute dissolved in 1000 mL (1 L) of solution - Molarity (units: molar = M).   So a 1 molar (1 M) solution contains 1 mole of solute in 1 L of solution. Concentration in M = no. moles / Vol in L Dr Toby Hudson

Chem 1001 Lecture 14 Reactions in Solution Q: A saline solution used in pharmaceuticals contains 0.10 M sodium chloride. What mass is needed to prepare 10 L of saline?   A: C = 0.1 M, V = 10. L n = CV = 0.10 mol L-1  10. L = 1.0 mol m = n  M = 1.0 mol  58.5 g mol-1 = 58.5 g Dr Toby Hudson

Chem 1001 Lecture 14 Reactions in Solution Q: A 3% w/v bleach solution contains 3.00 g sodium hypochlorite (NaOCl) in 100 mL of solution. What is its molarity?   A: Molar mass of NaOCl = 23.0 + 16.0 + 35.5 = 74.5 g mol-1 n = m/M = 3.00 g / 74.5 g mol-1 = 0.0403 mol C = n/V = 0.0403 mol / 0.1 L = 0.403 M Dr Toby Hudson

Chem 1001 Lecture 14 Reactions in Solution Q: What is the concentration of NaOH solution if a 25.00 mL portion of it reacts exactly with 22.40 mL of 0.104 M hydrochloric acid?   A: NaOH + HCl  NaCl + H2O V (NaOH) = 25.00 mL = 0.02500 L V (HCl) = 22.40 mL = 0.02240 L [HCl] = 0.104 M n (HCl used) = V x C = 0.02240 L  0.104 mol L-1 = 0.002330 mol   Dr Toby Hudson

Reactions in Solution A: NaOH + HCl  NaCl + H2O Chem 1001 Lecture 14 Reactions in Solution A: NaOH + HCl  NaCl + H2O   1 mol of HCl reacts with 1 mol of NaOH   \ 0.002330 mol of HCl reacts with 0.002330 mol of NaOH \ n (NaOH) = 0.002330 mol V (NaOH) = 25.00 mL = 0.02500 L \ [NaOH] = n/V = 0.002330 mol / 0.02500 L = 0.09318 M   Dr Toby Hudson

Learning Outcomes: By the end of this lecture, you should: know the meaning of the terms: solute, solvent and solution understand that molarity represents a concentration in moles per litre of solution be able to calculate concentrations given the volume and number of moles or mass of substance be able to use concentration and volume to determine amount of substance present in moles be able to complete the worksheet (if you haven’t already done so…)

Questions to complete for next lecture: Chem 1001 Lecture 4 Questions to complete for next lecture: Consider dissolving 23.4 g of sodium sulfate (Na2SO4) in enough water to form 125 mL of solution. Write the equation representing the dissolving of sodium sulfate. What is the molar mass of sodium sulfate? How many moles are there in 23.4 g of sodium sulfate? What is [Na2SO4]? What is [Na+] in this solution? What is [SO42-] in this solution?