What is the OH^- ion concentration of a solut | vedclass

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

Question

(B) Ammonium hydroxide $(NH_4OH)$ is a weak base and ammonium chloride $(NH_4Cl)$ is its salt with a strong acid,forming a basic buffer solution.For a

Vedclass · Accepted Answer

$9.0 	imes 10^{-4}$

Vedclass · Answer

(B) Ammonium hydroxide $(NH_4OH)$ is a weak base and ammonium chloride $(NH_4Cl)$ is its salt with a strong acid,forming a basic buffer solution.For a basic buffer,the concentration of $OH^-$ ions is given by the formula: $[OH^-] = K_b 	imes \frac{[Base]}{[Salt]}$Given: $K_b = 1.8 	imes 10^{-5}$,$[Base] = 0.05 \, M$,$[Salt] = 0.001 \, M$.Substituting the values: $[OH^-] = 1.8 	imes 10^{-5} 	imes \frac{0.05}{0.001}$$[OH^-] = 1.8 	imes 10^{-5} 	imes 50$$[OH^-] = 90 	imes 10^{-5} = 9.0 	imes 10^{-4} \, M$.

What is the $OH^-$ ion concentration of a solution containing $0.05 \, M$ ammonium hydroxide and $0.001 \, M$ ammonium chloride? Given $K_b(NH_4OH) = 1.8 \times 10^{-5}$.

Similar Questions

What will be the change in $pH$ by adding $0.1 \ M$ $CH_3COONa$ to $0.1 \ M$ $CH_3COOH$ at $298 \ K$ temperature? (Given: $pK_a$ of $CH_3COOH = 4.74$)

The $pH$ of a buffer solution containing $0.2 \ mol/L$ $CH_3COONa$ and $1.5 \ mol/L$ $CH_3COOH$ is ($K_a$ for acetic acid is $1.8 \times 10^{-5}$).

$A$ buffer solution contains equal concentrations of weak acid and its salt with a strong base. Calculate the $pH$ of the buffer solution if the dissociation constant of the weak acid is $1.8 \times 10^{-5}$.

The $pK_a$ of $HCN$ is $9.30$. What is the $pH$ of a solution prepared by mixing $2.5 \ mol$ of $KCN$ and $2.5 \ mol$ of $HCN$ in $500 \ mL$ of water (in $.30$)?

Out of the following,which pair of solutions is not a buffer solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module