What is the pH of a solution containing 7 g | vedclass

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

Question

(A) $1$. Calculate the molarity $(M)$ of the solution: $M = \frac{	ext{mass}}{	ext{molar mass} 	imes 	ext{volume in L}} = \frac{7 \ g}{35 \ g \ mo

Vedclass · Accepted Answer

$11.43$

Vedclass · Answer

(A) $1$. Calculate the molarity $(M)$ of the solution: $M = \frac{	ext{mass}}{	ext{molar mass} 	imes 	ext{volume in L}} = \frac{7 \ g}{35 \ g \ mol^{-1} 	imes 0.5 \ L} = 0.4 \ M$. $2$. For a weak base,$[OH^-] = \sqrt{K_b 	imes C} = \sqrt{1.8 	imes 10^{-5} 	imes 0.4} = \sqrt{0.72 	imes 10^{-5}} = \sqrt{7.2 	imes 10^{-6}} \approx 2.68 	imes 10^{-3} \ M$. $3$. Calculate $pOH$: $pOH = -\log[OH^-] = -\log(2.68 	imes 10^{-3}) = 3 - \log(2.68) \approx 3 - 0.43 = 2.57$. $4$. Calculate $pH$: $pH = 14 - pOH = 14 - 2.57 = 11.43$.

What is the $pH$ of a solution containing $7 \ g$ of $NH_4OH$ per $500 \ mL$? (Given: $K_b$ of $NH_4OH = 1.8 \times 10^{-5}$,Molar mass of $NH_4OH = 35 \ g \ mol^{-1}$)

Similar Questions

At $25^{\circ} C$,the percentage of ionization of '$x$' $M$ acetic acid is $4.242$. What is the value of $x$? $(K_a = 1.8 \times 10^{-5})$

The $pH$ of $0.01 \ M$ solution of acetic acid is $5.0$. What are the values of $[H^{+}]$ and $K_a$ respectively?

The ionization constant of acetic acid is $1.74 \times 10^{-5}$. Calculate the degree of dissociation of acetic acid in its $0.05 \ M$ solution. Calculate the concentration of acetate ion in the solution and its $pH$.

$A$ monobasic acid is $5 \%$ dissociated in its $0.02 \ M$ solution. Calculate the dissociation constant of the acid.

The $pH$ of $0.005 \, M$ codeine $(C_{18}H_{21}NO_{3})$ solution is $9.95$. Calculate its ionization constant and $pK_{b}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module