What is the [OH^-] in the final solution prep | vedclass

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

Question

(B) Step $1$: Calculate the millimoles of $H^+$ and $OH^-$.$n(H^+) = M 	imes V 	imes 	ext{basicity} = 0.050 \ M 	imes 20.0 \ mL 	imes 1 = 1.0 \ m

Vedclass · Accepted Answer

$0.10$

Vedclass · Answer

(B) Step $1$: Calculate the millimoles of $H^+$ and $OH^-$.$n(H^+) = M 	imes V 	imes 	ext{basicity} = 0.050 \ M 	imes 20.0 \ mL 	imes 1 = 1.0 \ mmol$.$n(OH^-) = M 	imes V 	imes 	ext{acidity} = 0.10 \ M 	imes 30.0 \ mL 	imes 2 = 6.0 \ mmol$.Step $2$: Calculate the remaining $OH^-$ after neutralization.$n(OH^-)_{	ext{remaining}} = 6.0 \ mmol - 1.0 \ mmol = 5.0 \ mmol$.Step $3$: Calculate the final concentration of $[OH^-]$.Total volume $= 20.0 \ mL + 30.0 \ mL = 50.0 \ mL$.$[OH^-] = \frac{5.0 \ mmol}{50.0 \ mL} = 0.10 \ M$.

What is the $[OH^-]$ in the final solution prepared by mixing $20.0 \ mL$ of $0.050 \ M \ HCl$ with $30.0 \ mL$ of $0.10 \ M \ Ba(OH)_2$?

Similar Questions

The hydrogen ion concentration in a mixture of $10 \ mL$ of $0.1 \ M$ $H_2SO_4$ and $10 \ mL$ of $0.1 \ M$ $KOH$ solution in water,is . . . . . . $M$.

What is the degree of dissociation of $CH_3COOH$ if $\wedge^{\circ}(CH_3COO^{-}) = 50 \ S \ cm^2 \ mol^{-1}$,$\wedge^{\circ}(H^{+}) = 350 \ S \ cm^2 \ mol^{-1}$ and the molar conductivity of $5 \times 10^{-2} \ M \ CH_3COOH$ is $20 \ S \ cm^2 \ mol^{-1}$?

The correct order of increasing $[H_3O^{+}]$ in the following aqueous solutions is:

The ionization constant of $ClCH_{2}COOH$ is $1.35 \times 10^{-3}$. What will be the $pH$ of $0.1 \ M$ acid and its $0.1 \ M$ sodium salt solution?

At $25^{\circ} C$, the ionic product of water is $10^{-14}$. The free energy change for the self-ionization of water in $kCal \ mol^{-1}$ is close to

Vedclass Test Series

Exam Paper Generator

Online Exam Module