Calculate the pH of a 0.10 M ammonia solution . Calculate the pH after 50.0 mL of this solution is treated with 25.0 mL of 0.10 M HCl. The dissociation constant of ammonia, `K_b=1.77xx10^(-5)`

Explanation

Calculation of pH of 0.1 M ammonia : Ammonia is a weak base initially `[NH_3]`= 0.1 M
In it x M is ionized So, At equilibrium .
`[NH_3]=(0.1 -x) M=0.1` M
`[NH_4^+]=[OH^-]`= x M
`{:(,NH_(3(aq)) + H_2O_((l)) hArr, NH_(4(aq))^(+)+, OH_((aq))^(-)),("At equilibrium :", (0.1-x), x M, x M),(, =0.1 M,,) :}`
`K_b=([NH_4^+][OH^-])/0.1 = x^2/0.1`
`therefore x=sqrt(0.1xxK_b)`
`=sqrt(0.1xx1.77xx10^(-5))`
`=sqrt(1.77xx10^(-6))`
`=1.3304xx10^(-3) M =[OH^-]`
pOH=-log `[OH^-]`
=-log `(1.3304xx10^(-3))`
=-(-2.876)=2.876
pH = 14.0-pOH=14.0-2.876 = 11.124
pH of mix solution after neutralization :
Base = (50 ml 0.1 M `NH_3`)
So mol of Base = `0.1 xx 50 / 1000 =5/1000`
Acid =(25 mL 0.1 M HCl)
So, mol of Acid = `0.1 xx 25/1000=2.5/1000`
After Neutralization 25 mL `NH_3`= 75 mL
In neutralization , Acid is consumed and base is remaining,
(After Neutralisation) = (Initial mol of Acid ) - (Initial mol of base )
`=(5/1000-2.5/1000)=2.5/1000` mol
Molarity of remaining = `"mol"/"Total volume"`
`=(2.5xx1000)/(1000xx75)`=0.0333 M
Here, Total volume =(50+25) = 75 mL = 75/1000 L
`NH_3 + H_2O hArr NH_4OH`
`{:(,NH_(3(aq))+H_2O_((l)) hArr ,NH_(4(aq))^(+) +, OH_((aq))^(-)),("Initial:", 0.0333 M, 0.0333, 0.0),("Molarity at equilibrium",(0.0333-y),+y,yM),(,approx 0.0333 ,=0.0333 , ):}`
`K_b=([NH_4^+][OH^-])/([NH_4OH])`
`1.77xx10^(-5)=((0.033)[OH^-])/(0.033)`
`therefore [OH^-]=1.77xx10^(-5)`
Calculation of `[H^+]` :
`[H^+]=K_w/([OH^-])=(1.0xx10^(-14))/(1.77xx10^(-5))`
`=5.6497xx10^(-10)`
Calculation of pH:
`pH=-log [H^+]=-log (5.6497xx10^(-10))`
=9.2480=9.25