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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^(+)]) :} |
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Answer» `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. |
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