Calculate degree of hydrolysis `K_(h)` and `pH` of `1M` urea hydtochlo – InterviewSolution

InterviewSolution

1.	Calculate degree of hydrolysis `K_(h)` and `pH` of `1M` urea hydtochloride solution in water `K_(b)` (Urea) `=1.5xx10^(-14)` at` 25^(@)C`.Consider urea as a monoacidic base.Take `log0.55=-0.26`
Answer» `NH_(2)CONH_(3)Cl`is a salt of `SA+WB` `K_(h)=(K_(W))/(K_(b))=(10^(-14))/(1.5xx10^(-14))=6.667xx10^(-1)` Now `h=sqrt((K_(h))/(C))=sqrt((10^(-14))/(1.5xx10^(-14)xx1))` or `h=0.816(gt0.1)` so we use actual relation `K_(h)=(Ch^(2))/(1-h)=(1)/(1.5)` `1.5h^(2)+h-1=0 rArr h=0.55` `[H^(+)]=ch=0.55M` `:.pH=0.26`

Discussion

No Comment Found

Related InterviewSolutions

Assertion: Addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate `AgBr` rather than `AgCl`. Reason : `K_(sp)` of `AgClltK_(sp)` of `AgBr`.A. Both A and R true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true
100ml of 0.1M `H_(3)PO_(4)` is titrated with 0.05 M NaOH solution till 2nd equivalent point. Then, resultant solution was mixed with 10 ml of 0.5 MHCI solution `K_(1)=10^(-3)M` `K_(2)=10^(-8)M` `K_(3)=10^(-13)M` Solubility of compound `A(OH)_(2)` in final solution will be :A. `10_(10)M`B. `4xx10^(-10)M`C. `4xx10^(-22)M`D. `4xx10^(-14)M`
Calculate degree of hydrolysis `K_(h)` and `pH` of `1M` urea hydtochloride solution in water `K_(b)` (Urea) `=1.5xx10^(-14)` at` 25^(@)C`.Consider urea as a monoacidic base.Take `log0.55=-0.26`
Calculate `[H^(+)]` in a `0.20M` solution of dichloriacetic acid `(K_(a)=5xx10^(-2))` that also contains `0.1M` sodium dichloroacetate.Neglect hydrolysis of sodium salt.