Problem Set - Normality and pH

JWP(2024) FREE for Students – NOT FOR RESALE Answer Key 4. H 2 SO 4 → 2H + + SO 4 2- Normality of H 2 SO 4 = 0.025M H 2 SO 4 × (2 eq. H + /1 mole H 2 SO 4 ) = 0.050N or 0.050 eq. H + / L Note: The term “eq.” stands for equivalent Ca(OH) 2 → 2OH - + Ca 2+ Normality of Ca(OH) 2 = 0.025M Ca(OH) 2 × (2 eq. OH - /1 mole Ca(OH) 2 ) = 0.050N or 0.050 eq. OH - / L Note: The term “eq.” stands for equivalent 5. 0.024N Mg(OH) 2 Molarity = 0.024N × (1 mole Mg(OH) 2 / 2 equivalent OH - ) = 0.012M Mg(OH) 2 0.024N implies 0.024M OH - ; pOH = -log [OH - ] = -log (0.024) = 1.62; pH = 12.38 6. 0.1000M × (2 equivalents of H + / 1 mol.) = 0.2000N 7. We need to calculate [H + ] before we can determine the pH We have the following dissociation equation based on the assumption that both protons dissociate completely in water in H 2 SO 4 H 2 SO 4 → 2H + + SO 4 2- We can write out the conversion factors based on the chemical equation shown above. 0.025N H 2 SO 4 = 0.025 eq. of H + / 1 L eq. refers to equivalent [H + ] = 0.025N H 2 SO 4 × (1 mole H 2 SO 4 /2 eq. H + ) × (2 mole H + /1 mole H 2 SO 4 ) [H + ] = 0.025M or 0.025 mole of H + /L As you can see, the unit of normality is implicitly equal to the molarity of H + in the solution. pH = -log [H+] = -log (0.025M) = 1.60 (2 significant figures only AFTER the decimal place)