Answer:
[tex]pH=4.63[/tex]
Explanation:
Hello!
In this case, since the ionization of ammonia, which is a weak base, is written as:
[tex]NH_3+H_2O\rightleftharpoons NH_4^++OH^-[/tex]
We can see that the ammonium ion is the conjugate acid whereas the hydroxide ions the conjugate base; that is why we use the Henderson-Hasselbach equation to compute the pH, given the pKb of ammonia 4.75:
[tex]pH=pKb+log(\frac{[conj\ acid]}{[base]} )[/tex]
In such a way, for the given moles of ammonia, base, and those of ammonium chloride, conjugate acid form, we obtain:
[tex]pH=4.75+log(\frac{0.15}{0.2} )\\\\pH=4.63[/tex]
Best regards!
