Equation is as under: pOH = pKb + log ([salt]/[base]) Now: We have to find pKb for NH3: As, pKb = - log [Kb] pKb of NH3 = -log Kb = -log (1.8*10^-5) = 4.74 Volume of final solution = 200+250 = 450mL As the equation of molarity is M1V1 = M2V2 So, M1V1 = M2V2 V1 = 450mL V2 = 250mL M2 = 0.15 M Putting all values into equation: M1*450 = 0.15*250 M1 = 0.15*250/450 M1 = 0.0833M
Now we have to find molarity of [NH3]: [NH3] M1V1 = M2V2 V1 = 50 mL V2 = 200mL M2 = 0.12M Putting all the values again in equation: M1*50 = 0.12*200 M1 = 0.12*200/450 M1 = 0.05333 Using the equation pOH = pKb + log ([salt]/[base]) pkb = 4.74 salt =0.0833 M base = 0.0533M Substitute: pOH = 4.74 + log(0.0833/0.0533) pOH = 4.74 + log 1.563 pOH = 4.74 + 0.19 pOH = 4.93
