Answer:
[tex]pH=9.32[/tex]
Explanation:
Hello there!
In this case, according to the given scenario, it turns out necessary for us to realize that the addition of NaOH consumes the conjugate base, NH4Cl, and produces more base, NH3; that is why the reaction taking place is:
[tex]NH_4^++OH^-\rightarrow NH_3+H_2O[/tex]
Thus, the reacting moles of ammonium and hydroxide ions are:
[tex]n_{NH_4^+}=0.0800L*0.188mol/L=0.01504mol\\\\n_{OH^-}=0.0200L*0.100mol/L=0.00200mol[/tex]
Thus, the resulting moles of ammonium and ammonia are respectively:
[tex]n_{NH_4^+}=0.01504mol-0.00200mol=0.01304mol\\\\n_{OH^-}=0.0800L*0.163mol/L+0.00200mol=0.01504mol[/tex]
Then, by using the Henderson-Hasselbach equation and recalling the pKb of ammonia (4.74), we first calculate the pOH as follows:
[tex]pOH=pKb+log(\frac{n_{NH_4^+}}{n_{NH_3}} )\\\\pOH=4.74+log(\frac{0.01304mol}{0.01504mol} )\\\\pOH=4.68[/tex]
And finally the pH:
[tex]pH=14-4.68\\\\pH=9.32[/tex]
Regards!
