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# UNCW CHM 101 - Electrolytes and Concentrations of Solutions

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CHM 101 1st Edition Lecture 10Lecture OverviewI. Acids and BasesII. Neutralization ReactionsLecture 9I. Identifying ElectrolytesII. Concentrations of solutionsIII. MolarityIV. DilutionsV. pH and pOH scaleIdentifying Electrolytes - use strong and weak acids and bases to predict strong and weak electrolytes - Ionic Compounds o Strong electrolytes – strong baseso Weak electrolytes – noneo Non-electrolytes – none- Molecular Compounds o Strong electrolytes – all strong acidso Weak electrolytes – weak acids and weak bases o Non-electrolytes – all othersConcentrations of Solutions- Concentration is very important- Concentration is expressed in molarity (M)o Molarity = moles of solute/liters of solution- Example: There is 25.0 g of KNO3 in 250 mL of water. What is the concentration of the solution in molarity (mol/L)o Find the formula weight of KNO3 101.11 g/molo 25.0 g KNO3 x 1 mol KNO3= 0.247 mol KNO3 101.11 g KNO3KNO3 = 0.247 mol KNO3/0.250 L = 0.989 M- Molarity can be used as a conversion factor in stoichiometry calculationso Example: What is the concentration of sodium in 3.5 M Na2CO3 solution? Na+ = 3.5 mol Na2CO3x 2 mol Na+ = 7.0 ML 1 mol NaCO3Dilutions - Solutions are often prepared in the lab by diluting small volumes of concentrated stock solutions to a larger volumeo Mconc x Vconc = Mdilx Vdil Conc – concentrated solution Dil – diluted solutiono Always make sure volume is in the same units on both sideso When doing these problems make sure you balance the equations first (like always) and write down what you know pH Scale- Measure of the concentration of the H+ ions in a solutiono pH less than 7 is acidic o pH greater than 7 is basico 7 is neutral - How can pure water have a pH?o A very small portion of the water molecules in an aqueous solution will dissociateinto H+ and OH- ions- pH is a negative log scale of H+ concentrationo pH = -logH+ o H+ = 10-pH Mo Example: what is the pH of a solution of 0.0065 M HNO3? pH = -log(0.0065) = pH 2.2- pOHo pOH = -logOH- o OH- = 10-pOH- pH + pOH = 14- Example: What is the pH of a 0.025 M solution of KOH?o KOH is a strong base o KOH (s)  K+ (aq) + OH- (aq) Base is OH- pOH = -log(0.025) = 1.6 pH = 14 – 1.6 =

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