Calculate the degree of ionisation and ph of 0.05 M solution of a weak base having the ionization constant (Kb) is1.77 ? 10 raised to power -5 .Also calculate the ionisation constant of the conjugate acid of this base

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Now, we know thatKa×Kb=Kw=10-14Therefore, Ka =KwKb=10-141.77×10-5=0.564×10-9Now, Ka=2or, α=KaC=0.564×10-90.05=10.92×10-9=1.044×10-4Now, B(aq) +H2O(aq)  BH+(aq) + OH-(aq)          0.05M                            0                  0         0.05-α                        +α               +αTherfore,[OH-]=1.044×10-4pOH=-log[OH-] =-log(1.044×10-4) =-0.0187+4=3.981So, pH= 14-pOH= 14-3.981=10.01

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