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

Explanation

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
Calculation of `[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