Question

Assuming complete ionization, which will have maximum pH?

• A. $0.01M$ ${NH}_{4}Cl$
• B. $0.01M$ ${\left({NH}_4\right)}_2{SO}_4$
• C. $0.01M$ ${\left({NH}_4\right)}_3{PO}_4$
• D. Equal

Answer 1

The correct option is C $0.01M$ ${\left({NH}_4\right)}_3{PO}_4$

$N{H}_{4}Cl$ is salt of $N{H}_{4}OH$ and $HCl$ so it dissociates to give only one $H^{+}$ ion.

$(N{H}_4{)}_{2}S{O}_4$ is salt of $N{H}_{4}OH$ and $H_{2}S{O}_4$ so it dissociates to give only two $H^{+}$ ions.

$(N{H}_4{)}_{3}P{O}_4$ is salt of $N{H}_{4}OH$ and $H_{3}P{O}_{4}Cl$ so it dissociates to give only three $H^{+}$ ion. So it has maximum $pH$.