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 $0.01 \ M \ BOH$ solution is $10$. What is its degree of dissociation (in $\%$)? (Given $K_{b}$ of $BOH$ is $1 \times 10^{-6}$)

The $K_a$ of propanoic acid is $1.4 \times 10^{-5}$. Calculate the $pH$ of a $0.1 \ M$ solution.

For two acids $A$ and $B$,the values of $pK_{a1} = 1.2$ and $pK_{a2} = 2.8$ are given. Which of the following is correct?

The $pH$ of a $0.1 \ M$ solution of acetic acid will be (degree of dissociation of acetic acid is $0.0132$).

Which of the following formulae is used to find the $[OH^{-}]$ ion concentration of a weak monoacidic base?

Vedclass Test Series

Exam Paper Generator

Online Exam Module