What is [NH_(4)^(+)] in a solution that contain 0.02 M NH_(3)(K_(b)=1.8xx10^(-5)) and 0.01 M KOH?
Calculate pH of 0.1M HC1.
Calculate pH of a resultant solution of 0.1 M HA (K_(a)=10^(-6)) and 0.45 M HB (K_(a)=2xx10^(-6)) at 25^(@)C.
Calculate the pH of 0.005 M NaOH solution.
Calculate pOH of 0.1 M aq. Solution of weak base BOH (K_(b)=10^(-7)) at 25^(@)C.
The [H^(+)] of a resulting solution that is 0.01 M acetic acid (K_(a)=1.8xx10^(-5)) and 0.01 M in benzoic acid (K_(a)=6.3xx10^(-5)) :
Solution of a weak acid and its anion (that is,its conjugate base) or of a base and its common cation are buffered. When we add a small amount of acid or base to any one of the, the pH of solution change very little. pH of buffer solution can be computed as for acidic buffer : pH=pK_(a)+ log.(["Conjugate base"])/(["Acid"]) for basic buffer : pOH=pK_(b)+log.(["Conjugate acid"])/([Base]) It is generly accepted that a has useful buffer cpacity (pH change resisting power) provided that the value of [salt or conjugate base] /[acid] for acidic buffer lies within the range of 1 : 10 to 1. Buffer capacity is maximum when [conjugate base] = [acid] Calculater the pH of a solution made by adding 0.01 mole of HCl in 100 mL of a solution which is 0.2 M in NH_(3)(pK_(b)=4.74) and 0.3 M in NH_(4)^(+) : (Assuming no change in volume )
What concentration of HCOO^(-) is present in a solution of weak of 0.01 M HCOOH (K_(a)=1.8xx10^(-4) and 0.01 M HCl?
An aqueous solution at room temperature contains 0.1 M NH_(4)Cl and 0.01M NH_(4)OH (pK_(b)=5), the pH of the solution is :
Calculate the pH of the followings : 0.001 M Ba(OH)_(2)
NARENDRA AWASTHI-IONIC EEQUILIBRIUM-Assertin-Reason Type Questions
