75 mL of 0.2 M HCl is mixed with 25 mL of | vedclass

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

Question

(A) Step $1$: Calculate the total moles of $HCl$ in the mixture.$n_1 = M_1 	imes V_1 = 0.2 \ M 	imes 75 \ mL = 15 \ mmol$$n_2 = M_2 	imes V_2 = 1 \

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Step $1$: Calculate the total moles of $HCl$ in the mixture.$n_1 = M_1 	imes V_1 = 0.2 \ M 	imes 75 \ mL = 15 \ mmol$$n_2 = M_2 	imes V_2 = 1 \ M 	imes 25 \ mL = 25 \ mmol$Total moles of $HCl = n_1 + n_2 = 15 + 25 = 40 \ mmol$.Step $2$: Calculate the total volume of the final solution.$V_{total} = V_1 + V_2 + V_{water} = 75 \ mL + 25 \ mL + 300 \ mL = 400 \ mL$.Step $3$: Calculate the final molarity $(M_{final})$ of $HCl$.$M_{final} = \frac{	ext{Total moles}}{	ext{Total volume}} = \frac{40 \ mmol}{400 \ mL} = 0.1 \ M$.Step $4$: Calculate the $pH$ of the solution.Since $HCl$ is a strong acid,$[H^+] = [HCl] = 0.1 \ M = 10^{-1} \ M$.$pH = -\log_{10}[H^+] = -\log_{10}(10^{-1}) = 1$.

$75 \ mL$ of $0.2 \ M \ HCl$ is mixed with $25 \ mL$ of $1 \ M \ HCl$. To this solution,$300 \ mL$ of distilled water is added. What is the $pH$ of the resultant solution?

Similar Questions

What is the $pH$ of a solution in which $25.0 \ mL$ of $0.1 \ M \ NaOH$ is added to $25 \ mL$ of $0.08 \ M \ HCl$ and the final solution is diluted to $500 \ mL$?

$A$ base dissolved in water yields a solution with a hydroxyl ion concentration of $0.05 \ mol \ L^{-1}$. The solution is

At $298 \, K$,what is the concentration of hydrogen ions in a $10^{-8} \, M \, HCl$ aqueous solution? $(K_W = 10^{-14})$

By adding $20 \ mL$ of $0.1 \ N \ HCl$ to $20 \ mL$ of $0.001 \ N \ KOH$,the $pH$ of the resulting solution will be:

Which statement is true for a solution of $0.020 \ M \ H_2SO_4$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module