B2.(b)Calculate the pH of a basic buffer made by dissolving 0.125 moles of NHCls and 0.25 moles of NH)in 500 mL of water.Ko for NH)is 1.8 x 10 pK=14-pKb=14-(log1o(K)=14-(log101.8x10=9.25 [acid][NH"(ag)]=0.125 mol /0.5 L=0.25 M [base]=[NH3()=0.25 mol /0.5 L=0.50 M pH=k-loans明-925-log.25-955 B3.Calculate the standard enthalpy of formation of strontium carbonate,SrCOs given the following data: Srg+%02e→Sr0g △H°=-592 kJ mol SrC03s→SrOs+C02gl △H°=+234 kJmol C(onpe)+02g→C0e AH=-394 kJ mol The reaction whose enthalpy we want is S+C+1.This reaction is obtained by adding the above reactions as follows: Srs+hO2g→SrOs △H°=-592 kJ mol" C(gaphite))+O2g)→CO2g △H°=-394 kJmol SrOs+CO2g→SrCO3s △H°=-(+234kmo) Srg+C(graphite)+1.502g→SrC03g△H°=（←592-394-234）=-1220 kJmolB2. (b) Calculate the pH of a basic buffer made by dissolving 0.125 moles of NH4Cl(s) and 0.25 moles of NH3(aq) in 500 mL of water. Kb for NH3(aq) is 1.8 x 10-5. pKa = 14 – pKb = 14 – (-log10(Ka) = 14 – (-log10(1.8 x 10-5) = 9.25 [acid] = [NH4 + (aq)] = 0.125 mol / 0.5 L = 0.25 M [base] = [NH3(aq)] = 0.25 mol / 0.5 L = 0.50 M a 10 10 [acid] .25 pH pK log 9.25 log 9.55 [base] .5 =− = − = B3. Calculate the standard enthalpy of formation of strontium carbonate, SrCO3(s) given the following data: Sr(s) + ½ O2(g) → SrO(s) ΔHo = −592 kJ mol-1 SrCO3(s) → SrO(s) + CO2(g) ΔHo = +234 kJ mol-1 C(graphite) + O2(g) → CO2(g) ΔHo = −394 kJ mol-1 The reaction whose enthalpy we want is Sr(s) + C(graphite) + 1.5 O2(g) → SrCO3(s). This reaction is obtained by adding the above reactions as follows: Sr(s) + ½ O2(g) → SrO(s) ΔHo = −592 kJ mol-1 C(graphite) + O2(g) → CO2(g) ΔHo = −394 kJ mol-1 SrO(s) + CO2(g) → SrCO3(s) ΔHo = − (+234 kJ mol-1) ________________________________________________ Sr(s) + C(graphite) + 1.5 O2(g) → SrCO3(s) ΔHo = (– 592 – 394 – 234) = – 1220 kJ mol-1 4