By adding 20 mL 0.1 N HCl to 20 mL 0.1 N | vedclass

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

Question

(B) The reaction between $HCl$ (a strong acid) and $KOH$ (a strong base) is a neutralization reaction: $HCl + KOH \rightarrow KCl + H_2O$. Since the n

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) The reaction between $HCl$ (a strong acid) and $KOH$ (a strong base) is a neutralization reaction: $HCl + KOH ightarrow KCl + H_2O$. Since the number of equivalents of $HCl$ $(20 \ mL 	imes 0.1 \ N = 2 \ mEq)$ is equal to the number of equivalents of $KOH$ $(20 \ mL 	imes 0.1 \ N = 2 \ mEq)$,the reaction results in complete neutralization. The resulting salt,$KCl$,is formed from a strong acid and a strong base,which does not undergo hydrolysis. Therefore,the solution remains neutral,and the $pH = 7$.

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

Similar Questions

$A$ $20 \ mL$ sample of a $0.2 \ M$ solution of the weak diprotic acid $H_2A$ is titrated with $0.25 \ M \ NaOH$. Molarity of the solution at the second equivalence point is:

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

The incorrect relation for isohydric solutions is $:-$

One mole of ammonia was completely absorbed in one litre solution each of $(a)$ $1 \ M \ HCl$,$(b)$ $1 \ M \ CH_{3}COOH$,and $(c)$ $1 \ M \ H_{2}SO_{4}$ at $298 \ K$. The decreasing order for the $pH$ of the resulting solutions is (Given $pK_{b}(NH_{3}) = 4.74$)

Which of the following absorbs oxygen rapidly?

Vedclass Test Series

Exam Paper Generator

Online Exam Module