Calculate the pH of an aqueous solution of 1.0 M ammonium formate assuming complete dissociation (pKa of formic acid = 3.8 and pKb of ammonia = 4.8)

Question

Calculate the pH of an aqueous solution of 1.0 M ammonium formate assuming complete dissociation (pKa of formic acid = 3.8 and pKb of ammonia = 4.8) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For salts of weak acid and weak base pH = 7 + (1/2) ( pKa - pKb) = 7 + (1/2) ( 3.8 - 4.8) = 6.50

Q. Calculate the pH of an aqueous solution of $1.0 M$ ammonium formate assuming complete dissociation
$(p K_{a}$ of formic acid $= 3.8$ and $p K_{b}$ of ammonia $= 4.8$ )

For salts of weak acid and weak base
$p H = 7 + \frac{1}{2} (p K_{a} - p K_{b}) = 7 + \frac{1}{2} (3.8 - 4.8) = 6.50$

Q. Calculate the pH of an aqueous solution of 1.0M ammonium formate assuming complete dissociation (pKa​ of formic acid =3.8 and pKb​ of ammonia =4.8)

For salts of weak acid and weak base pH=7+21​(pKa​−pKb​)=7+21​(3.8−4.8)=6.50

Q. Calculate the pH of an aqueous solution of $1.0 M$ ammonium formate assuming complete dissociation
$(p K_{a}$ of formic acid $= 3.8$ and $p K_{b}$ of ammonia $= 4.8$ )

For salts of weak acid and weak base
$p H = 7 + \frac{1}{2} (p K_{a} - p K_{b}) = 7 + \frac{1}{2} (3.8 - 4.8) = 6.50$