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Calculate the pH of a 0.10 M ammonia solution . Calculate the pH after 50.0 mL of this solutionis treated with 25.0 mL of 0.10 M HCl. The dissociation constant of ammonia, K_b=1.77xx10^(-5) |
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Answer» Solution :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 molof 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 |
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