20 mL solution contains 0.1 M NH_4Cl and 0 | vedclass

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

Question

(A) The given solution is a basic buffer consisting of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$.According to the Henderson-Ha

Vedclass · Accepted Answer

Addition of $1 \ mL$ water

Vedclass · Answer

(A) The given solution is a basic buffer consisting of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$.According to the Henderson-Hasselbalch equation for a basic buffer,$pOH = pK_b + \log \frac{[Salt]}{[Base]}$.Dilution of a buffer solution with water does not change the ratio of the concentration of the salt to the concentration of the base,as both concentrations decrease by the same factor.Therefore,the $pH$ of a buffer solution remains unchanged upon the addition of a small amount of water.

$20 \ mL$ solution contains $0.1 \ M \ NH_4Cl$ and $0.01 \ M \ NH_4OH$. By adding which one will its $pH$ not change?

Similar Questions

$A$ buffer solution contains $0.1 \ mol$ of sodium acetate dissolved in $1000 \ cm^{3}$ of $0.1 \ M$ acetic acid. To the above buffer solution,$0.1 \ mol$ of sodium acetate is further added and dissolved. The $pH$ of the resulting buffer is

An acidic buffer is obtained on mixing

Find the concentration of sodium acetate when added to $0.1 \ M$ solution of acetic acid to form a buffer solution of $pH = 5.5$ (in $M$)? $(pK_{a} \text{ of } CH_{3}COOH = 4.5)$

The $pOH$ of a basic buffer (e.g.,$NH_4OH/NH_4Cl$) is $5$. If the concentration of the salt is tripled while the concentration of the base remains constant,find the new $pOH$ value. (Given: $\log 3 = 0.48$)

Assertion : Mixture of $CH_3COOH$ and $CH_3COONH_4$ is an example of acidic buffer.
Reason : Acidic buffer contains equimolar mixture of a weak acid and its salt with weak base.

Vedclass Test Series

Exam Paper Generator

Online Exam Module