200 mL of 0,1M NH4OH are mixed with 200mL of 0,1M NH4Cl. What is the pH of the final solution? #K_b=1,8*10^{-5}# (The answer given by the authors pH=9,3)

1 Answer
anor277
Jul 11, 2018

Now the buffer equation holds that....

Explanation:

#pH=pK_a + log_10{(["Conjugate base"])/(["Conjugate acid"])}#..

And in this scenario...

#pH=pK_a + log_10([[NH_3]]/[[NH_4^+]])#..

(and note here that since the concentrations of ammonium ion, and ammonia are EQUAL, their logarithm of their RATIO is EQUAL to ZERO..i.e. #log_10(a/a)=log_(10)1=0#...because #10^0=1#)...

But we are quoted #pK_b# NOT #pK_a#...but we know that...

#14=pK_a+pK_b#...#pK_a=14-log_10(1.80xx10^-5)=14-4.75=9.26#...

And hence the #pH# of the buffer is #~=9.3# AS REQUIRED.....

Note here that so-called ammonium hydroxide, #NH_4OH# is actually aqueous ammonia, #NH_3•H_2O#, ammonium hydroxide is very much a misnomer.

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