Use log 1.8=0.26, `k_a` of formic acid =`1.8xx10^(-4),k_a` of acetic acid =`1.8xx10^(-5),k_b` of ammonia=`1.8xx10^(4)k_(a1)` of `H_2S=10^(-7)` and `k_
Use log 1.8=0.26, `k_a` of formic acid =`1.8xx10^(-4),k_a` of acetic acid =`1.8xx10^(-5),k_b` of ammonia=`1.8xx10^(4)k_(a1)` of `H_2S=10^(-7)` and `k_(a2)` of `H_2S=10^(-14)`, for the following matchings
Match the entries of column II for which the equality or inequality given in the column I are satisfied.
`{:("Column-I","Column-II"),((A)10^(-5)M HCl "solution" gt0.1 M H_2S "solution",(p)alpha_("water")("degree of dissociation of water")),((B)CH_3COOH"solution at pH equal to 4.74"=NH_4OH "solution at pH equal to 9.26",(q)[OH^(-)]),((C )0.1 M CH_3COOH "solution"=1.0 M HCOOH solution,(r)alpha("degree of dissociation of electrolytes")),((D)"0.1 M of a weak acid" HA_1(k_a=10^(-5))"solution" lt "0.01 M of a weak acid" HA_2(k_a=10^(-6))"solution",(s)pH),(,(t)1/[H^(+)]) :}`
1 Answers
Correct Answer - A-p,q,r,s,t ; B-p,r ; C-r ; D-p,q,s,t
(A)`10^(-5)` M HCl solution gt 0.1 M `H_2S` solution
`alpha` of HCl is more as compared to `H_2S` but concentration of `H_2S` is high.
So `[H^(+)]` ion form HCl is =`10^(-5)` M and `[H^+]` ion from `H_2S=sqrt(K_1C)=sqrt(10^(-7)xx0.1)=10^(-4) M`.
Hence `[OH^-]` ion for HCl =`10^(-9)` and `[OH^-]` coming from HCl is low hence common ion effect is low.
(B)pH of `CH_3COOH` =4.74 and pOH of `NH_4OH=4.74`
Hence `[H^+]` ion from `CH_3COOH` =4.74 and pOH of `NH_4OH`=4.74.
Hence `[H^+]` ion from `CH_3COOH` and `[OH^-]` ion from `NH_4OH` is same.
(C )`alpha=sqrt((Ka)/C)` : for `CH_3COOH,alpha=sqrt(1.8xx10^(-4))`
for `HCOOH, alpha=sqrt(1.8xx10^(-4))`
So, answer is r.
(D)`[H^+]` ion from weak acid `HA_1=sqrt(C_1Ka_1)=sqrt(0.1xx10^(-5))=10^(-3)`
`[H^+]` ion from weak acid `HA_2=sqrt(C_2Ka_2)=sqrt(0.01xx10^(-6))=10^(-4)`
So, answer is p,q,s.