The pH of a 0.1 M solution of NH_{4}OH (havi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For a weak base,the concentration of hydroxyl ions is given by $[OH^{-}] = \sqrt{K_{b} 	imes C}$.$[OH^{-}] = \sqrt{1.0 	imes 10^{-5} 	imes 0.1}

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) For a weak base,the concentration of hydroxyl ions is given by $[OH^{-}] = \sqrt{K_{b} 	imes C}$.$[OH^{-}] = \sqrt{1.0 	imes 10^{-5} 	imes 0.1} = \sqrt{1.0 	imes 10^{-6}} = 1.0 	imes 10^{-3} \ M$.Using the ionic product of water,$K_{w} = [H^{+}] [OH^{-}] = 1.0 	imes 10^{-14}$.$[H^{+}] = \frac{K_{w}}{[OH^{-}]} = \frac{1.0 	imes 10^{-14}}{1.0 	imes 10^{-3}} = 1.0 	imes 10^{-11} \ M$.The $pH$ is calculated as $pH = -\log [H^{+}]$.$pH = -\log (1.0 	imes 10^{-11}) = 11$.

The $pH$ of a $0.1 \ M$ solution of $NH_{4}OH$ (having $K_{b} = 1.0 \times 10^{-5}$) is equal to

Similar Questions

The $pH$ of a $0.01 \, M$ solution of acetic acid having a degree of dissociation of $12.5 \%$ is:

Sulphurous acid $(H_{2}SO_{3})$ has $Ka_{1} = 1.7 \times 10^{-2}$ and $Ka_{2} = 6.4 \times 10^{-8}$. The $pH$ of $0.588 \ M \ H_{2}SO_{3}$ is ..... . (Round off to the Nearest Integer)

Dimethyl amine $(CH_3)_2NH$ is a weak base and its ionization constant is $5.4 \times 10^{-5}$. Calculate $[OH^{-}]$,$[H_3O^{+}]$,$pOH$ and $pH$ of its $0.2 \ M$ solution at equilibrium.

The $pH$ of $0.001 \ M$ acetic acid is $.......$

The percentage dissociation of a decinormal solution of a weak acid $HA$ is: $(K_a = 4.9 \times 10^{-8})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module