This problem is providing us with the concentration of NH4Cl and NaOH and asks for the resulting pH as they are mixed. At the end, the result turns out to be 5.299.
In chemistry, one can calculate the pH when two substances are mixed by firstly evaluating the relevant equilibrium reaction, which omits every ion derived from strong bases or acids.
Thus, for the equilibrium in this problem, one first write the reaction between ammonium chloride and sodium hydroxide as:
[tex]NaOH+NH_4Cl\rightarrow NH_4OH+NaCl[/tex]
Thus, assuming equal volumes of NH4Cl and NaOH (1 L each), one can calculate how much ammonium chloride is left:
[tex]n_{NH_4Cl}=0.12mol-0.03mol=0.09mol[/tex]
And its concentration:
[tex][NH_4Cl]=\frac{0.09mol}{1L+1L}=0.045M[/tex]
Next, we write the equilibrium between the ammonium ions and water, as chloride ions are not said to form any HCl as it is an strong acid:
[tex]NH_4^++H_2O\rightleftharpoons NH_3+H_3O^+[/tex]
Whose equilibrium expression can be written as:
[tex]Ka=\frac{[NH_3][H_3O^+]}{[NH_4^+]} \\\\10^{-9.25}=\frac{x*x}{0.045M-x}\\ \\5.62x10^{-10}=\frac{x^2}{0.045M}[/tex]
Next, we solve for x with the low-Ka approximation:
[tex]x=\sqrt{ 5.62x10^{-10}*0.045M}=5.03x10^{-6}M[/tex]
Which also equals the concentration of hydronium ions. Thereby, the pH turns out to be:
[tex]pH=-log(5.03x10^{-6})=5.299[/tex]
Learn more about pH calculations: https://brainly.com/question/1195974
