Calculate the pH of the solution in which 0.2 M NH_4Cl and 0.1 M NH_3 are present . The pK_b of ammonia solution is pK_b= 4.75

Calculate the pH of the solution in which 0.2 M NH_4Cl and 0.1 M NH_3

1.	Calculate the pH of the solution in which 0.2 M NH_4Cl and 0.1 M NH_3 are present . The pK_b of ammonia solution is pK_b= 4.75
Answer» Solution :Calculation of `K_b` : `pK_b=-log (K_b)`=(4.75) `therefore log K_b=-4.75` `therefore K_b=10^(-4.75)=1.7783xx10^(-5)` `approx 1.78xx10^(-5)` Calculation of `[OH^-]` : `{:("Strong salt :", NH_4Cl to , NH_(4(AQ))^(+)+, CL^(-)),(,0.2,0.2M,0.2M):}` `{:("Weak base: ",NH_(3(aq))+ , H_2O_((l)) hArr , NH_(4(aq))^(+)+, OH_((aq))^(-)),("Equilibrium of " NH_3 " Initial (M):", 0.1 , -, +0,+0),("Change reaction :", -X M, ,+xM,+xM),("CONCENTRATION at equilibrium:",(0.1-x)M,,(0.2+x)M,xM) :}` `K_b=([NH_4^+][OH^-])/([NH_3])` `1.78xx10^(-5)=((0.2+x)(x))/((0.1-x))` x is neglected in additional substraction . Value of `K_b` is much less , the value of x is very less , Therefore x is NEGLIGIBLE in comparison to 0.2 and 0.1 . `therefore (0.2+x) approx 0.2` and `(0.1 -x) approx 0.1` `therefore 1.78xx10^(-5) =(0.2x)/0.1` `therefore x=[OH^-]=(1.78xx10^(-5)xx0.1)/0.2` `= 0.88xx10^(-5)` M Calculationof `[H^+]` : `[H^+] =K_w/([OH^-]) = (1xx10^(-14))/(0.88xx10^(-5))=1.136xx10^(-9)` Calculation of pH : pH=-log `[H^+] = log 1.136xx10^(-9) = 8.9445 approx` 8.95