20 mL of 0.5 N HCl and 35 mL of 0.1 N N | vedclass

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

Question

(D) Step $1$: Calculate the milliequivalents of $HCl$ and $NaOH$.$mEq 	ext{ of } HCl = 20 \ mL 	imes 0.5 \ N = 10 \ mEq$.$mEq 	ext{ of } NaOH = 35

Vedclass · Accepted Answer

Turn methyl orange red

Vedclass · Answer

(D) Step $1$: Calculate the milliequivalents of $HCl$ and $NaOH$.$mEq 	ext{ of } HCl = 20 \ mL 	imes 0.5 \ N = 10 \ mEq$.$mEq 	ext{ of } NaOH = 35 \ mL 	imes 0.1 \ N = 3.5 \ mEq$.Step $2$: Determine the nature of the resulting solution.Since $mEq 	ext{ of } HCl > mEq 	ext{ of } NaOH$,the solution is acidic.$mEq 	ext{ of } H^+ 	ext{ remaining} = 10 - 3.5 = 6.5 \ mEq$.Step $3$: Analyze the indicators.Phenolphthalein is colorless in acidic medium and pink in basic medium. Methyl orange is red in acidic medium $(pH  4.4)$.Calculating the final concentration of $H^+$:$[H^+] = \frac{6.5 \ mEq}{55 \ mL} \approx 0.118 \ N$.$pH = -\log(0.118) \approx 0.93$.At $pH \approx 0.93$,methyl orange turns red. Therefore,the correct option is $D$.

$20 \ mL$ of $0.5 \ N \ HCl$ and $35 \ mL$ of $0.1 \ N \ NaOH$ are mixed. The resulting solution will

Similar Questions

What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

When dilute $H_2SO_4$ is added to an aqueous solution of potassium chromate,the yellow color of the solution changes to orange. This indicates that .....

The concentration of $[H^{+}]$ and concentration of $[OH^{-}]$ of a $0.1 \ M$ aqueous solution of $2\%$ ionised weak acid is [Ionic product of water $= 1 \times 10^{-14}$]

$5 \ mL$ of $0.4 \ N$ $NaOH$ is mixed with $20 \ mL$ of $0.1 \ N$ $HCl$. The $pH$ of the resulting solution will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module